smc-4268-50-Determining Spontaneity

CHEMISTRY, M4 EQ-Bank 17

The decomposition of a metal carbonate is represented by the following equation:

$\ce{MCO3(s) → MO(s) + CO2(g)}$

The following data was recorded:

$\Delta H = +130 \, \text{kJ/mol},\ \ \Delta S = +160 \, \text{J/mol K}$

Calculate the Gibbs free energy at 350 K. (2 marks)

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Determine if the reaction is spontaneous at this temperature. (1 mark)

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Discuss how both enthalpy and entropy influence the spontaneity of this reaction and predict the temperature range in which the reaction will be spontaneous. (4 marks)

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a. $\Delta G = +74 \, \text{kJ/mol}$

b. The reaction is non-spontaneous at 350 K.

c. Enthalpy and entropy influence the spontaneity:

→ To be spontaneous, $\Delta G$ must be negative, where $\Delta G = \Delta H- T\Delta S$.

→ The reaction has a positive enthalpy, indicating it requires energy input, which opposes spontaneity.

→ However, the positive entropy means the disorder of the system increases, which favours spontaneity.

→ At higher temperatures, the entropy contribution will dominate, overcoming the positive enthalpy and making the reaction spontaneous.

→ To find the temperature where the reaction becomes spontaneous, set $\Delta G = 0$:

$0=\Delta H-T\Delta S$

$T=\dfrac{\Delta H}{\Delta S}=\dfrac{130}{0.160}=812.5\ \text{K}$

→ Thus, the reaction will be spontaneous above 812.5 K.

Show Worked Solution

a. $\Delta G = \Delta H- T\Delta S$

Convert $\Delta S$ to $\text{kJ/mol K}$

$\Delta S = 0.160 \, \text{kJ/mol K}$

$\Delta G = 130-(350 \times 0.160) = 130-56 = +74 \, \text{kJ/mol}$

b. Since $\Delta G > 0$, the reaction is non-spontaneous at 350 K.

c. Enthalpy and entropy influence the spontaneity:

→ To be spontaneous, $\Delta G$ must be negative, where $\Delta G = \Delta H- T\Delta S$.

→ The reaction has a positive enthalpy, indicating it requires energy input, which opposes spontaneity.

→ However, the positive entropy means the disorder of the system increases, which favours spontaneity.

→ At higher temperatures, the entropy contribution will dominate, overcoming the positive enthalpy and making the reaction spontaneous.

→ To find the temperature where the reaction becomes spontaneous, set $\Delta G = 0$:

$0=\Delta H-T\Delta S$

$T=\dfrac{\Delta H}{\Delta S}=\dfrac{130}{0.160}=812.5\ \text{K}$

→ Thus, the reaction will be spontaneous above 812.5 K.

CHEMISTRY, M4 EQ-Bank 16

Given the following values for a reaction:

$\Delta H = +150 \, \text{kJ/mol}$ and $\Delta S = +250 \, \text{J/mol K}$

Calculate the Gibbs free energy at 298 K. (2 mark)
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Is the reaction spontaneous at this temperature? Justify your answer. (1 mark)

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Predict the temperature at which the reaction would become spontaneous. (2 marks)

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a. $\Delta G = +75.5 \, \text{kJ/mol}$

b. The reaction is non-spontaneous at 298 K.

c. The reaction becomes spontaneous above 600 K.

Show Worked Solution

a. $\Delta G = \Delta H-T\Delta S$:

Convert $\Delta S$ to $\text{kJ/mol K}$:

$\Delta S = 0.250 \, \text{kJ/mol K}$

$\Delta G = 150-(298 \times 0.250) = 150-74.5 = +75.5 \, \text{kJ/mol}$

b. $\Delta G > 0$ at 298 K $\Rightarrow$ reaction is non-spontaneous at 298 K.

c. Find the temperature where the reaction becomes spontaneous:

→ Solve for $T$ when $\Delta G = 0$:

$0= \Delta H-T\Delta S$

$T = \dfrac{\Delta H}{\Delta S} = \dfrac{150}{0.250} = 600 \, \text{K}$

→ As the temperature rises, $T\Delta S$ increases, therefore $\Delta G$ decreases, increasing the spontaneity of the reaction.

→ The reaction becomes spontaneous above 600 K.

CHEMISTRY, M4 EQ-Bank 7 MC

At what temperature would the reaction with the following enthalpy and entropy values become non-spontaneous?

$\Delta H = +100 \, \text{kJ/mol}$ and $\Delta S = +200 \, \text{J/mol K}$

Below 500 K
Above 500 K
Below 250 K
Above 250 K

Show Answers Only

$A$

Show Worked Solution

→ To determine when the reaction becomes non-spontaneous, solve for $T$ when $\Delta G = 0$:

$0 = \Delta H-T\Delta S$

$T = \dfrac{\Delta H}{\Delta S} = \dfrac{100}{0.200} = 500 \, \text{K}$

→ As the temperature increases, $T\Delta S$ increases and so $\Delta G$ decreases. For a non-spontaneous reaction, $\Delta G$ must be postive.

→ Thus, the reaction becomes non-spontaneous below 500 K.

$\Rightarrow A$

CHEMISTRY, M4 EQ-Bank 6 MC

The chemical reaction has the following enthalpy and entropy values:

$\Delta H = -400 \, \text{kJ/mol}$ and $\Delta S = +150 \, \text{J/mol K}$

Determine the spontaneity of the reaction.

Always spontaneous
Spontaneous at low temperatures only
Spontaneous at high temperatures only
Non-spontaneous at all temperatures

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$A$

Show Worked Solution

→ For a reaction to be spontaneous, $\Delta G$ must be negative.

→ Using $\Delta G = \Delta H-T\Delta S$, since $\Delta H$ is negative and $\Delta S$ is positive, $\Delta G$ will always be negative, making the reaction spontaneous at all temperatures.

$\Rightarrow A$

CHEMISTRY, M4 EQ-Bank 15

A wide range of chemical reactions take place in the combustion chamber of car engines. One such reaction occurs when carbon monoxide is converted to carbon dioxide:

$\ce{2CO(g) + O2(g) \rightarrow 2CO2(g)}$

The following values were obtained under standard conditions:

$\Delta H = +283 \, \text{kJ/mol}, \quad \Delta S = +86.6 \, \text{J/mol K}$

Calculate the Gibbs free energy for the above reaction at 298 K. (2 marks)

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Explain the effect of enthalpy and entropy on the spontaneity of this reaction. (3 marks)

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a. $\Delta G = +257.2 \, \text{kJ/mol}$

b. Effect of enthalpy and entropy on spontaneity:

→ Since the entropy value for this reaction is positive, the reaction is considered to be entropy-driven as this will drive towards a negative value for the Gibbs free energy and contribute to the reaction being spontaneous.

→ However, the enthalpy value of this reaction is also positive, which does not contribute towards a negative value for the Gibbs free energy and in turn contributes towards making this reaction non-spontaneous.

→ Since the magnitude of the enthalpy value is greater, the entropy drive is unable to overcome this and results in a non-spontaneous reaction overall.

Show Worked Solution

a. Using the Gibbs free energy equation:

$\Delta G = \Delta H-T \Delta S$

Convert $\Delta S$ to $\text{kJ/mol K}$:

$\Delta S = +0.0866 \, \text{kJ/mol K}$.

Now, substitute the values into the equation:

$\Delta G = 283 \, \text{kJ/mol}-(298 \, \text{K} \times 0.0866 \, \text{kJ/mol K}) = 283-25.8 = +257.2 \, \text{kJ/mol}$

$\Rightarrow$ Since $\Delta G > 0$, the reaction is non-spontaneous.

b. Effect of enthalpy and entropy on spontaneity:

→ Since the magnitude of the enthalpy value is greater, the entropy drive is unable to overcome this and results in a non-spontaneous reaction overall.

CHEMISTRY, M4 EQ-Bank 5 MC

The combustion of butane occurs via an exothermic reaction.

$\ce{2C4H10(g) + 13 O2(g) \rightarrow 8 CO2(g) + 10 H2O(g)}$

Which of the following correctly describes the spontaneity of this reaction?

Spontaneous at all temperatures
Spontaneous at low temperatures
Spontaneous at high temperatures
Non-spontaneous at all temperatures

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$A$

Show Worked Solution

→ This reaction is exothermic ($\Delta H < 0$) and also exhibits an increase in entropy ($\Delta S > 0$) due to the production of gaseous products from smaller hydrocarbon molecules.

→ As both $\Delta H$ and $\Delta S$ are favourable, the reaction will be spontaneous at all temperatures according to the Gibbs free energy equation:

$\Delta G = \Delta H-T \Delta S$

→ Since $\Delta H$ is negative and $\Delta S$ is positive, $\Delta G$ will always be negative, making the reaction spontaneous at all temperatures.

$\Rightarrow A$

CHEMISTRY, M4 EQ-Bank 14

The combustion of propane is represented by the following equation:

$\ce{C3H8(g) + 5O2(g) \rightarrow 3CO2(g) + 4H2O(l)}$

$\Delta H = -2220\ \text{kJ mol}^{-1}$, $\Delta S = -269\ \text{J mol}^{-1}\text{K}^{-1}$

Justify why the entropy change for this reaction is negative based on the chemical equation. (2 marks)

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Using the Gibbs free energy equation, determine if the combustion of propane is spontaneous at 298 K. Show your working. (2 marks)

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a. Entropy change in the combustion of propane:

→ The reactants include 1 molecule of propane gas and 5 molecules of oxygen gas, making a total of 6 gas molecules.

→ The products, however, consist of 3 molecules of carbon dioxide gas and 4 molecules of liquid water.

→ Since gases have higher entropy than liquids, the total number of gas molecules decreases from 6 to 3, which results in a decrease in disorder.

→ This explains why the entropy change, $\Delta S$, is negative.

b. Using the Gibbs free energy equation:

$\Delta G = \Delta H-T \Delta S$

Convert $\Delta S$ from $\text{J mol}^{-1}\text{K}^{-1}$ to $\text{kJ mol}^{-1}\text{K}^{-1}$:

$\Delta S = -0.269\ \text{kJ mol}^{-1}\text{K}^{-1}$.

$\Delta G$	$= -2220\ \text{kJ mol}^{-1}-(298\ \text{K} \times -0.269\ \text{kJ mol}^{-1}\text{K}^{-1})$
	$= -2220\ \text{kJ mol}^{-1} + 80.162\ \text{kJ mol}^{-1}$
	$= -2139.84\ \text{kJ mol}^{-1}$

$\Rightarrow$ Since $\Delta G < 0$, the reaction is spontaneous at 298 K.

Show Worked Solution

a. Entropy change in the combustion of propane:

→ The reactants include 1 molecule of propane gas and 5 molecules of oxygen gas, making a total of 6 gas molecules.

→ The products, however, consist of 3 molecules of carbon dioxide gas and 4 molecules of liquid water.

→ Since gases have higher entropy than liquids, the total number of gas molecules decreases from 6 to 3, which results in a decrease in disorder.

→ This explains why the entropy change, $\Delta S$, is negative.

b. Using the Gibbs free energy equation:

$\Delta G = \Delta H-T \Delta S$

Convert $\Delta S$ from $\text{J mol}^{-1}\text{K}^{-1}$ to $\text{kJ mol}^{-1}\text{K}^{-1}$:

$\Delta S = -0.269\ \text{kJ mol}^{-1}\text{K}^{-1}$.

$\Delta G$	$= -2220\ \text{kJ mol}^{-1}-(298\ \text{K} \times -0.269\ \text{kJ mol}^{-1}\text{K}^{-1})$
	$= -2220\ \text{kJ mol}^{-1} + 80.162\ \text{kJ mol}^{-1}$
	$= -2139.84\ \text{kJ mol}^{-1}$

$\Rightarrow$ Since $\Delta G < 0$, the reaction is spontaneous at 298 K.

CHEMISTRY, M4 EQ-Bank 3 MC

Given the relationship $\Delta G = \Delta H-T\Delta S$, which of the following statements is correct?

A reaction is always spontaneous if $\Delta S$ is negative and $\Delta H$ is negative.
A reaction is always spontaneous if $\Delta S$ is negative and $\Delta H$ is positive.
A reaction is always spontaneous if $\Delta S$ is positive and $\Delta H$ is negative.
A reaction is always spontaneous if $\Delta S$ is positive and $\Delta H$ is positive.

Show Answers Only

$C$

Show Worked Solution

→ A reaction is spontaneous when $\Delta G < 0$.

→ For this to always happen, $\Delta S$ must be positive (increasing disorder), and $\Delta H$ must be negative (exothermic).

→ Therefore, $\Delta G$ will be negative at all temperatures.

CHEMISTRY, M4 EQ-Bank 13

The reduction of iron (III) oxide takes place as follows:

$\ce{Fe2O3(s) + 3H2(g) \rightarrow 2Fe(s) + 3H2O(l)}$

$ \Delta H = 98.8\ \text{kJ mol}^{-1}$, $\Delta S = 141\ \text{J mol}^{-1}\text{K}^{-1}$

Predict the temperature at which this reaction will become spontaneous. Show your working. (2 marks)

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$700.7\ \text{K}$

Show Worked Solution

Using the Gibbs free energy equation:

$\Delta G = \Delta H-T \Delta S$

For spontaneity, $\Delta G < 0$ , so we solve for $T$ when $\Delta G = 0$

$0 = \Delta H-T \Delta S\ \Rightarrow \ T = \dfrac{\Delta H}{\Delta S}$

$\Rightarrow$ Convert $\Delta S$ from $\text{J mol}^{-1}\text{K}^{-1}$ to $\text{kJ mol}^{-1}\text{K}^{-1}$:

$T = \dfrac{98.8\ \text{kJ mol}^{-1}}{0.141\ \text{kJ mol}^{-1}\text{K}^{-1}} = 700.7\ \text{K}$

$\Rightarrow$ The reaction becomes spontaneous above approximately 700.7 $\text{K}$

CHEMISTRY, M4 EQ-Bank 10

Calculate the temperature range where the following reaction will occur spontaneously.

$\ce{2H2O2(l) -> 2H2O(l) + O2(g)}$

where $\Delta H = -196.0\ \text{kJ}$ and $\Delta S = 125.0\ \text{J K}^{-1} $ (3 marks)

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→ The reaction is spontaneous at temperatures below 1568.0 K.

Show Worked Solution

$\text{Find}\ T\ \text{when}\ \Delta G = 0: $

$\Delta H-T \Delta S =0$

$T= \dfrac{\Delta H}{\Delta S}= \dfrac{-196.0}{0.125}=1568.0\ \text{K} $

→ As $T$ decreases, $\Delta G$ decreases.

→ Reaction is spontaneous when $\Delta G < 0$.

→ The reaction is spontaneous at temperatures below 1568.0 K.

CHEMISTRY, M4 EQ-Bank 9

At what temperature range would the following reaction occur spontaneously?

$\ce{C(s) + O2(g) -> CO2(g)}$

where $\Delta H = -393.5\ \text{kJ}$ and $\Delta S = 213.6\ \text{J K}^{-1} $ (3 marks)

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→ The reaction is spontaneous at all temperatures.

Show Worked Solution

$\text{Find}\ T\ \text{when}\ \Delta G = 0: $

$\Delta H-T \Delta S$$=0$

$T= \dfrac{\Delta H}{\Delta S} = \dfrac{-393.5}{0.2136}= -1842.52\ \text{K}$

→ Since negative temperature is not physically meaningful in this context, the reaction is spontaneous at all temperatures.

→ As $T$ increases, $\Delta G$ becomes more negative.

→ Reaction is spontaneous when $\Delta G < 0$.

→ The reaction is spontaneous at all temperatures.

CHEMISTRY, M4 EQ-Bank 8

Determine the temperature range at which the following reaction will occur spontaneously.

$\ce{N2(g) + 3H2(g) -> 2NH3(g)}$

where $\Delta H = -92.4\ \text{kJ}$ and $\Delta S = -198.3\ \text{J K}^{-1} $ (3 marks)

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→ The reaction is spontaneous at temperatures below 465.94 K.

Show Worked Solution

$\text{Find}\ T\ \text{when}\ \Delta G = 0: $

$\Delta H-T \Delta S$$=0$

$T= \dfrac{\Delta H}{\Delta S} = \dfrac{-92.4}{-0.1983} =465.94\ \text{K} $

→ As $T$ decreases, $\Delta G$ decreases.

→ Reaction is spontaneous when $\Delta G < 0$.

→ Therefore the reaction is spontaneous at temperatures below 465.94 K.

CHEMISTRY, M4 EQ-Bank 7

At what temperature range would the following reaction occur spontaneously?

$\ce{2SO2(g) + O2(g) -> 2SO3(g)}$

where $\Delta H = -198.2\ \text{kJ}$ and $\Delta S = -188\ \text{J K}^{-1} $ (3 marks)

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→ The reaction is spontaneous at temperatures below 1053.19 K.

Show Worked Solution

$\text{Find}\ T\ \text{when}\ \Delta G = 0: $

$\Delta H-T \Delta S$$=0$

$T= \dfrac{\Delta H}{\Delta S}= \dfrac{-198.2}{-0.188} =1053.19\ \text{K} $

→ As $T$ decreases, $\Delta G$ decreases.

→ Reaction is spontaneous when $\Delta G < 0$.

→ The reaction is spontaneous at temperatures below 1053.19 K.

CHEMISTRY, M4 EQ-Bank 11

Consider the reaction below at 298 K:

$\ce{N2(g) + 3H2(g) -> 2NH3(g)}$

The standard enthalpies of formation and standard entropies are as follows:

$\begin{aligned}
\ce{N2(g)}: & \quad \Delta H_f^\circ = 0 \, \text{kJmol}^{-1}, & S^\circ = 191.5 \, \text{J/mol·K} \\
\ce{H2(g)}: & \quad \Delta H_f^\circ = 0 \, \text{kJmol}^{-1}, & S^\circ = 130.7 \, \text{J/mol·K} \\
\ce{NH3(g)}: & \quad \Delta H_f^\circ = -45.9 \, \text{kJmol}^{-1}, & S^\circ = 192.3 \, \text{J/mol·K} \\
\end{aligned}$

Calculate the Gibbs free energy ($ \Delta G $) for the reaction and determine whether the reaction is spontaneous at 298 K. Show all relevant calculations. (4 marks)

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$\Delta G =-32.7\ \text{kJ/mol} $, indicating the reaction is spontaneous at 298 K.

Show Worked Solution

To calculate the Gibbs free energy ($ \Delta G $) for the reaction, we use the following equation:

$ \Delta G = \Delta H-T \Delta S $

First, calculate the change in enthalpy ($ \Delta H $) for the reaction:

$ \Delta H_{\text{reaction}}^\circ = \sum \Delta H_f^\circ (\text{products}) -\ \sum \Delta H_f^\circ (\text{reactants})$

$ \Delta H_{\text{reaction}}^\circ = [2 \times (-45.9)]-[1 \times (0) + 3 \times (0)] = -91.8 \, \text{kJ/mol}$

Calculate the change in entropy ($ \Delta S $) for the reaction:

$ \Delta S_{\text{reaction}}^\circ$	$= \sum S^\circ (\text{products})-\sum S^\circ (\text{reactants}) $
	$= (2 \times (192.3))-(1 \times (191.5) + 3 \times (130.7))$
	$= 384.6-(191.5 + 392.1)$
	$= -199.0\ \text{J/mol·K}$
	$=-0.199\ \text{kJ/mol·K}$

Calculate $ \Delta G $:

$\Delta G= \Delta H-T \Delta S= -91.8-(298 \times -0.199)= -32.5\ \text{kJ/mol}$

$\Rightarrow$ The reaction is spontaneous at 298 K.

CHEMISTRY, M4 EQ-Bank 12

When aqueous solutions of $\ce{Na^+}$ and $\ce{Cl^-}$ are mixed, the following exothermic reaction occurs:

$\ce{Na+(aq) + Cl-(aq) -> NaCl(aq)}$

If the temperature of the mixture is increased, explain how this will affect the Gibbs free energy and whether the reaction will become more or less spontaneous. In your response, refer to both the enthalpy and entropy changes involved in the reaction. (3 marks)

Show Answers Only

→ Gibbs free energy is calculated using $\Delta G = \Delta H\ \ -\ \ T\Delta S$.

→ Since the reaction is exothermic, the enthalpy change $\Delta H$ will be negative.

→ In this reaction, two ions combine to form a single ion, leading to a decrease in entropy $\Delta S$, making it negative.

→ As the temperature $T$ increases, the term $-T \Delta S$ becomes more positive, which in turn causes $\Delta G$ to increase.

→ As a result, the reaction becomes less spontaneous at higher temperatures.

Show Worked Solution

→ Gibbs free energy is calculated using $\Delta G = \Delta H\ \ -\ \ T\Delta S$.

→ Since the reaction is exothermic, the enthalpy change $\Delta H$ will be negative.

→ In this reaction, two ions combine to form a single ion, leading to a decrease in entropy $\Delta S$, making it negative.

→ As the temperature $T$ increases, the term $-T \Delta S$ becomes more positive, which in turn causes $\Delta G$ to increase.

→ As a result, the reaction becomes less spontaneous at higher temperatures.

CHEMISTRY, M4 EQ-Bank 5

The following reaction represents the conversion of diamond to graphite:

$\ce{2C_{diamond} \rightarrow 2C_{graphite}}$

- $\ce{\Delta $H$_{f}\ C_{diamond} = 1.9 kJ mol^{-1}}$
- $\ce{\Delta $S$_{f}\ C_{diamond} = 2.38 J mol^{-1} K^{-1}}$
- $\ce{\Delta $S$_{f}\ C_{graphite} = 5.74 J mol^{-1} K^{-1}}$

Determine $\Delta G$ at 298K and state whether the reaction is spontaneous or not. (3 marks)

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What does $\Delta G$ indicate about the rate of reaction? (1 mark)

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i. $\Delta G = -5.8025\ \text{kJ}$

→ Reaction is spontaneous.

ii. Rate of reaction and $\Delta G$:

→ Gibbs Free Energy doesn’t indicate anything about the kinetics of the reaction. While the conversion of diamond to graphite is spontaneous, it occurs over millions of year

Show Worked Solution

→ Standard enthalpy and entropy of elements in their natural state is 0.

$\Delta H$	$= \Sigma H_{\text{products}}-\Sigma H_{\text{reactants}}$
	$= (2 \times 0)-(2 \times 1.9)$
	$=-3.8\ \text{kJ mol}^{-1} $

$\Delta S$	$=\Sigma S_{\text{products}}-\Sigma S_{\text{reactants}}$
	$= (2 \times 5.74)-(2 \times 2.38)$
	$= 11.48-4.76 $
	$=6.72\ \text{J mol}^{-1}\ \text{K}^{-1}$

$\Delta G$	$= \Delta H-T\Delta S$
	$= -3.8-(298 \times 0.00672)$
	$= -5.8025\ \text{kJ}$

→ The reaction is spontaneous as $\Delta G < 0$.

ii. Rate of reaction and $\Delta G$:

→ Gibbs Free Energy doesn’t indicate anything about the kinetics of the reaction. While the conversion of diamond to graphite is spontaneous, it occurs over millions of year

CHEMISTRY, M4 EQ-Bank 3

A 3.1g sample of $\ce{CaCO3_{(s)}}$ decomposes into $\ce{CaO_{(s)}}$ and $\ce{CO2_{(g)}}$. Entropy values for these chemicals are given below and the molar enthalpy for the reaction is 360 kJ/mol.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex}\text{Substance}\rule[-1ex]{0pt}{0pt} & \text{Standard Entropy}\ (\Delta S) \\
\hline
\rule{0pt}{2.5ex}\ce{CaCO3}\rule[-1ex]{0pt}{0pt} & \text{92.88 J/K} \\
\hline
\rule{0pt}{2.5ex}\ce{CaO(s)}\rule[-1ex]{0pt}{0pt} & \text{39.75 J/K} \\
\hline
\rule{0pt}{2.5ex}\ce{CO2(g)}\rule[-1ex]{0pt}{0pt} & \text{213.6 J/K} \\
\hline
\end{array}

Write a chemical equation for the above reaction. (1 mark)

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Calculate the entropy change for this reaction. (2 marks)

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Determine whether the reaction is spontaneous or non-spontaneous at room temperature. (2 marks)

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$\ce{CaCO3(s) \rightarrow CaO(s) + CO2(g)}$
$\ce{4.97 J K^{-1}}$
$\text{Since}\ \Delta G > 0,\ \text{the reaction is not spontaneous.}$

Show Worked Solution

i. $\ce{CaCO3(s) \rightarrow CaO(s) + CO2(g)}$

ii.	$\Delta S$	$=\Sigma S_{\text{products}}-\Sigma S_{\text{reactants}}$
		$= 213.6 + 39.75-92.88$
		$= 160.47\ \text{J mol}^{-1}\ \text{K}^{-1}$

$\ce{n(CaCO3)}= \dfrac{\text{m}}{\text{MM}} = \dfrac{3.1}{100.09} = 0.03097\ \text{mol} $

$\text{Entropy change}\ = 160.47 \times 0.03097 = 4.97\ \text{J K}^{-1}$

iii. $\text{Room Temperature = 298.15 K}$

$\Delta G$	$=\Delta H-T \Delta S$
	$=360-(298.15 \times 0.16047) $
	$= 312.179\ \text{kJ}$

$\text{Since}\ \Delta G > 0,\ \text{the reaction is not spontaneous.}$

\(\Delta G\)	\(= -2220\ \text{kJ mol}^{-1}-(298\ \text{K} \times -0.269\ \text{kJ mol}^{-1}\text{K}^{-1})\)
	\(= -2220\ \text{kJ mol}^{-1} + 80.162\ \text{kJ mol}^{-1}\)
	\(= -2139.84\ \text{kJ mol}^{-1}\)

\(\Delta G\)	\(= -2220\ \text{kJ mol}^{-1}-(298\ \text{K} \times -0.269\ \text{kJ mol}^{-1}\text{K}^{-1})\)
	\(= -2220\ \text{kJ mol}^{-1} + 80.162\ \text{kJ mol}^{-1}\)
	\(= -2139.84\ \text{kJ mol}^{-1}\)

\( \Delta S_{\text{reaction}}^\circ\)	\(= \sum S^\circ (\text{products})-\sum S^\circ (\text{reactants}) \)
	\(= (2 \times (192.3))-(1 \times (191.5) + 3 \times (130.7))\)
	\(= 384.6-(191.5 + 392.1)\)
	\(= -199.0\ \text{J/mol·K}\)
	\(=-0.199\ \text{kJ/mol·K}\)

\(\Delta H\)	\(= \Sigma H_{\text{products}}-\Sigma H_{\text{reactants}}\)
	\(= (2 \times 0)-(2 \times 1.9)\)
	\(=-3.8\ \text{kJ mol}^{-1} \)

\(\Delta S\)	\(=\Sigma S_{\text{products}}-\Sigma S_{\text{reactants}}\)
	\(= (2 \times 5.74)-(2 \times 2.38)\)
	\(= 11.48-4.76 \)
	\(=6.72\ \text{J mol}^{-1}\ \text{K}^{-1}\)