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CHEMISTRY, M5 2024 HSC 30

An equilibrium mixture of hydrogen, carbon dioxide, water and carbon monoxide is in a closed, 1 L container at a fixed temperature as shown:

\(\ce{H2(g) +CO_2(g) \rightleftharpoons H2O(g) +CO(g)} \quad \quad K_{eq}=1.600\)

The initial concentrations are  \(\left[\ce{H2}\right]=1.000 \text{ mol L}^{-1}, \left[ \ce{CO2}\right]=0.500\ \text{mol L}^{-1},\ \left[\ce{H2O}\right]=0.400 \text{ mol L}^{-1}\)  and  \([ \ce{CO} ]=2.000 \text{ mol L} ^{-1}\).

An unknown amount of \(\ce{CO(g)}\) was added to the same container, and the temperature was kept constant. After the new equilibrium had been established, the concentration of \(\ce{H2O(g)}\) was found to be 0.200 mol L\(^{-1}\).

Using this information, calculate the unknown amount (in mol) of \(\ce{CO(g)}\) that was added to the container.   (4 marks)

Show Answers Only

\(4.92\ \text{mol}\)

Show Worked Solution
  • \(\ce{n_{intial}(CO(g))} = 0.400 \text{ and } \ce{n_{final}(CO(g))} = 0.200\).
  • \(\text{Change in the number of moles in } \ce{CO(g)} = 0.400-0.200 = 0.200\ \text{mol in } 1\ \text{L}\)

\begin{array} {|c|c|c|c|c|}
\hline & \ce{H2(g)} & \ce{CO2(g)} & \ce{H2O(g)} & \ce{CO2(g)} \\
\hline \text{Initial} & 1 & 0.5 & 0.4 & 2 + x \\
\hline \text{Change} & +0.2 & +0.2 & -0.2 & -0.2 \\
\hline \text{Equilibrium} & 1.2 & 0.7 & 0.2 & 1.8 + x \\
\hline \end{array}

  • Since all substances are present in a 1 L container, the concentrations of each substance is equal to the number of moles of that substance present at equilibrium
\(K_{eq}\) \(=\dfrac{\ce{[H2O(g)][CO(g)]}}{\ce{[H2(g)][CO2(g)]}}\)  
\(1.600\) \(=\dfrac{0.2 \times (1.8+x)}{1.2 \times 0.7}\)  
\(1.8 + x\) \(=1.6 \times \dfrac{1.2 \times 0.7}{0.2}\)  
\(x\) \(=6.72-1.8\)  
  \(=4.92\ \text{mol}\)  

 

  • 4.92 mol of \(\ce{CO(g)}\) were added to the container.
♦ Mean mark 55%.

CHEMISTRY, M5 2024 HSC 23

Consider the following equilibrium system.

\(\ce{\left[Co \left(H_2O\right)_6\right]^{2+}(aq) + 4Cl^{-} (aq) \rightleftharpoons\left[CoCl_4\right]^{2-}(aq) +6H_2O (l)}\)

\(\ce{\left[Co\left(H_2O\right)_6\right]^{2+}(aq)}\)  is pink and  \(\ce{\left[CoCl_4\right]^{2-}(aq)}\)  is blue. When a solution of these ions and chloride ions is heated, the mixture becomes more blue.

Relate the observed colour change to the change in \(K_{e q}\).   (3 marks)

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  • When the solution is heated and the mixture becomes more blue, it suggests that the concentration of \(\ce{\left[CoCl_4\right]^{2-}(aq)}\) is increasing.
  • The increase in temperature favoured the forward endothermic reaction and shifts the equilibrium position to the products.
  • Therefore the concentrations of \(\ce{\left[Co\left(H_2O\right)_6\right]^{2+}(aq)}\) and \(\ce{Cl^-(aq)}\) will decrease.
  • As  \(K_{eq}=\dfrac{\ce{\left[\left[CoCl4\right]^{2-}\right]}}{\ce{\bigl[\left[Co\left(H2O\right)6\right]^{2+}\bigr]\left[Cl^{-}\right]^4}}\),  \(K_{eq}\) will increase.
Show Worked Solution
  • When the solution is heated and the mixture becomes more blue, it suggests that the concentration of \(\ce{\left[CoCl_4\right]^{2-}(aq)}\) is increasing.
  • The increase in temperature favoured the forward endothermic reaction and shifts the equilibrium position to the products.
  • Therefore the concentrations of \(\ce{\left[Co\left(H_2O\right)_6\right]^{2+}(aq)}\) and \(\ce{Cl^-(aq)}\) will decrease.
  • As  \(K_{eq}=\dfrac{\ce{\left[\left[CoCl4\right]^{2-}\right]}}{\ce{\bigl[\left[Co\left(H2O\right)6\right]^{2+}\bigr]\left[Cl^{-}\right]^4}}\),  \(K_{eq}\) will increase.

CHEMISTRY, M5 2023 HSC 20 MC

Nitrogen monoxide and oxygen combine to form nitrogen dioxide, according to the following equation.

\( \ce{2NO(g) + O2(g) \rightleftharpoons 2NO2(g) \quad $K$_{e q}=2.47 \times 10^{12}} \)

A 2.00 L vessel is filled with 1.80 mol of \( \ce{NO2(g)} \) and the system is allowed to reach equilibrium.

What is the equilibrium concentration of \( \ce{NO(g)} \)?

  1. \( \text{0.00 mol L}^{-1}\)
  2. \( 4.34 \times 10^{-5}\  \text{mol L}^{-1} \)
  3. \(6.90 \times 10^{-5}\  \text{mol L}^{-1}\)
  4. \(8.69 \times10^{-5}\  \text{mol L}^{-1}\)
Show Answers Only

\(D\)

Show Worked Solution
  • As 1.80 mol of \( \ce{NO2(g)} \) is added to the solution, the reverse reaction can be used to determine the equilibrium concentration of \( \ce{NO(g)} \).
  •    \(\ce{2NO2(g) \rightleftharpoons 2NO(g) + O2(g)}\)
  • Reverse reaction  \(K_{eq} = \dfrac{\ce{[O_2][NO]^2}}{\ce{[NO_2]^2}}\)
  • Forward reaction  \(K_{eq}\) is the inverse of \(K_{eq}\) of the reverse reaction:
  •    \(K_{eq}=\dfrac{1}{2.47 \times 10^{12}}=4.0486 \times 10^{-13}\)

\begin{array} {|l|c|c|c|}
\hline  & \ce{2NO_2(g)} & \ce{2NO(g)} & \ce{O_2(g)} \\
\hline \text{Initial} & \ \ \ \ 0.9 & \ \ \ \ 0 & 0 \\
\hline \text{Change} & -2x & +2x & \ \ \ +x \\
\hline \text{Equilibrium} & \ \ \ \ 0.9 -2x & \ \ \ \ 2x & \ \ \ \ \ \ x \\
\hline \end{array}

  • \(-2x\) is very small as the \(K_{eq}\) for the reaction is very small, thus  \(0.9-2x \approx 0.9\).
  • By substituting the values into the \(K_{eq}\) for the reverse reaction: 
\(4.0486 \times 10^{-13}\) \(=\dfrac{(x)(2x)^2}{(0.9)^2}\)  
\(4.0486 \times 10^{-13}\) \(=\dfrac{4x^3}{(0.9)^2}\)  
\(4x^3\) \(=3.279 \times 10^{-13}\)  
\(x\) \(=4.344 \times 10^{-5}\)  

 

  • \(\ce{[NO2] = 2 \times 4.344 \times 10^{-5} = 8.69 \times 10^{-5}\ \text{mol L}^{-1}}\)

\(\Rightarrow D\)

♦♦ Mean mark 34%.

CHEMISTRY, M5 2023 HSC 37

When performing industrial reductions with \(\mathrm{CO}(\mathrm{g})\), the following equilibrium is of great importance.

\( \ce{2CO(g) \rightleftharpoons CO2(g) + C(s) \quad \quad $K$_{e q}  = 10.00  at 1095 K } \)

A 1.00 L sealed vessel at a temperature of 1095 K contains \( \ce{CO(g)} \) at a concentration of 1.10 × 10\(^{-2}\) mol L\(^{-1}\), \(\ce{CO2(g)} \) at a concentration of 1.21 × 10\(^{-3}\) mol L\(^{-1}\), and excess solid carbon.

  1. Is the system at equilibrium? Support your answer with calculations.   (2 marks)

  1. Carbon dioxide gas is added to the system above and the mixture comes to equilibrium. The equilibrium concentrations of \( \ce{CO(g)}\) and \(\ce{CO2(g)} \) are equal. Excess solid carbon is present and the temperature remains at 1095 K.

    Calculate the amount (in mol) of carbon dioxide added to the system.   (3 marks)

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a.    \(Q=\dfrac{\ce{[CO2]}}{\ce{[CO]^2}}=\dfrac{1.21 \times 10^{-3}}{(1.10 \times 10^{-2})^2}=10.0\)

\(\text{Since}\ \ Q=K_{eq},\ \text{system is in equilibrium.}\)
 

b.    \(0.143\ \text{mol} \)

Show Worked Solution

a.    \(Q=\dfrac{\ce{[CO2]}}{\ce{[CO]^2}}=\dfrac{1.21 \times 10^{-3}}{(1.10 \times 10^{-2})^2}=10.0\)

\(\text{Since}\ \ Q=K_{eq},\ \text{system is in equilibrium.}\)
 

b.    \(\ce{\text{Given}\ \ [CO]=[CO2]}, \)

\(K_{eq} =\dfrac{\ce{[CO2]}}{\ce{[CO]^2}} =\dfrac{1}{\ce{[CO]}} = 10.00\)

\(\Rightarrow \ce{[CO] = \dfrac{1}{10.00} = 0.1000 \text{mol L}^{-1}} \)

\(\Rightarrow \ce{[CO2] = 0.1000 \text{mol L}^{-1}} \)

From this point, the change in \(\ce{CO}\) and \(\ce{CO2}\) concentrations can be calculated…

♦♦♦ Mean mark (b) 24%.

\begin{array} {|l|c|c|c|}
\hline  & \ce{2CO(g)} & \ce{CO2(g)} & \ce{C(s)} \\
\hline \text{Initial} & 1.10 \times 10^{-2} &  1.21 \times 10^{-3} &  \\
\hline \text{Change} & +0.0890 & +0.0988 &  \\
\hline \text{Equilibrium} & \ \ \ 0.1000 & \ \ \ 0.1000 &  \\
\hline \end{array}

However, the change in moles of \(\ce{CO2}\) in the system consists of:

  • Change in \(\ce{CO2}\) concentration
  • Change in \(\ce{CO}\) concentration (as some of the added \(\ce{CO2}\) was converted into \(\ce{CO}\))

\(\ce{n(CO2)\ \text{required to increase}\ [CO] by 0.0988\ \text{mol}\ \ \ \text{(1 litre vessel)}}\)

\(\ce{\text{Formula ratio shows}\ \ CO2:CO = 1\ \text{mol} : 2\ \text{mol}} \)

\(\ce{n(CO2)\ \text{to add to increase}\ [CO2] = 0.0988\ \text{mol}\ \ \ \text{(1 litre vessel)}}\)

\(\ce{n(CO2)_{\text{total to add}} = 0.0988\ \text{mol} + n(CO2\ \text{to make CO)}} \)

\(\ce{n(CO2)\ \text{to add to increase}\ [CO] = \dfrac{0.0890}{2} = 0.0445\ \text{mol}}\)

\(\ce{n(CO2)_{\text{total to add}} = 0.0988 + 0.0445 = 0.143\ \text{mol}} \)

CHEMISTRY, M5 EQ-Bank 22

Hydrogen gas reacts with iodine gas to form hydrogen iodide according to the following equation.

\( \ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)}\) at 700 k

At equilibrium, the concentrations for \(\ce{H2, I2}\) and \(\ce{HI}\) are as follows:  0.326 mol L–1, 0.326 mol L–1 and 2.39 mol L–1 respectively.

What is the value of the equilibrium constant for this reaction?  (2 marks)

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53.7

Show Worked Solution

\[\ce{$K_{eq}$ = \frac{[HI]^2}{[H2][I2]} = \frac{2.39^2}{[0.326][0.326]} = 53.7}\]

CHEMISTRY, M5 EQ-Bank 3 MC

Consider the following reaction.

\( \ce{2NOCl(g) \rightleftharpoons 2NO(g) + Cl2(g)}\).

What is the equilibrium expression for this reaction?

  1. \(\dfrac{\ce{2[NO][Cl2]}}{\ce{2[NOCl]}}\)
  2. \(\dfrac{\ce{[NO]^2[Cl2]}}{\ce{[NOCl]^2}}\)
  3. \(\dfrac{\ce{2[NOCl]}}{\ce{2[NO][Cl2]}}\)
  4. \(\dfrac{\ce{[NOCl]^2}}{\ce{2[NO]^2[Cl2]}} \)
Show Answers Only

`B`

Show Worked Solution
  • All elements in the reaction are included in the \(\ce{$K_{eq}$}\) expression (i.e. no solids or pure liquids are present which would be omitted).
  •    \(\ce{$K_{eq}$} = \dfrac{\ce{[NO]^2[Cl2]}}{\ce{[NOCl]^2}}\)

\(\Rightarrow B\)

CHEMISTRY, M5 2019 HSC 31

The following reaction occurs in an aqueous solution.

   \(\ce{HgCl4^2-(aq) + Cu^2+(aq) \rightleftharpoons CuCl4^2-(aq) + Hg^2+\ \ \ \ \ \ $K_{eq}$ = 4.55 \times 10^{-11}}\)

A solution containing a mixture of \(\ce{HgCl4^2-(aq)}\) and \(\ce{Cu^2+(aq)}\) ions is prepared. The initial concentration of each ion is 0.100 mol L ¯1 and there are no other ions present.

Calculate the concentration of \(\ce{Hg^2+(aq)}\) ions once the system has reached equilibrium.   (4 marks)

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\(\ce{[Hg^2+] = 6.75 \times 10^{-7} mol L^{-1}}\)

Show Worked Solution

\[\ce{$K_{eq}$ = \frac{[CuCl4^2-][Hg^2+]}{[HgCl4^2-][Cu^2+]}}\]

\begin{array} {|l|c|c|c|c|}
\hline  & \ce{[HgCl4^2-]} & \ce{[Cu^2+]} & \ce{[CuCl4^2-]} & \ce{[Hg^2+]} \\
\hline \text{Initial} & 0.100 & 0.100 & 0 & 0 \\
\hline \text{Change} & -x & -x & +x & +x \\
\hline \text{Equilibrium} & 0.100-x & 0.100-x & x & x \\
\hline \end{array}

Since `x` is small  `=> 0.100-x~~0.100`

`4.55 xx 10^{-11}` `=(x xx x)/((0.100-x)(0.100-x))`  
`4.55 xx 10^{-11}` `=x^2/(0.100)^2`  
`x^2` `=4.55 xx 10^{-11} xx (0.100)^2`  
`x` `=sqrt(4.55 xx 10^{-11} xx (0.100)^2)`  
  `=6.75 xx 10^{-7}\ text{mol L}^{-1}`  

 
\(\therefore \ce{[Hg^2+] = 6.75 \times 10^{-7} mol L^{-1}}\)

CHEMISTRY, M6 2019 HSC 27

The relationship between the acid dissociation constant, `K_a`, and the corresponding conjugate base dissociation constant, `K_b`, is given by:

`K_(a)xxK_(b)=K_(w)`

Assume that the temperature for part (a) and part (b) is 25°C.

  1. The `K_a` of hypochlorous acid `text{(HOCl)}` is  `3.0 xx10^(-8)`.
  2. Show that the `K_b` of the hypochlorite ion, `text{OCl}^-`, is  `3.3 xx10^(-7)`.   (1 mark)

  3. The conjugate base dissociation constant, `K_b`, is the equilibrium constant for the following equation:
  4.      `text{OCl}^(-)(aq)+ text{H}_(2) text{O}(l) ⇌ text{HOCl}(aq)+ text{OH}^(-)(aq)`
  5. Calculate the pH of a 0.20 mol L¯1 solution of sodium hypochlorite `(text{NaOCl})`.   (4 mark)

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  1. `K_b=3.3 xx 10^{-7}`
  2. \(\ce{pH = 14-3.59 = 10.41}\)
Show Worked Solution

a.   `K_(a)xxK_(b)=K_(w)\ \ =>\ \ K_b=(K_(w))/(K_(a))`

`K_b=(1.0 xx 10^{-14})/(3.0 xx 10^{-8}=3.3 xx 10^{-7}`
 

b.   \(\ce{OCl-(aq) + H2O(l) \rightleftharpoons HOCl(aq) + OH-(aq)}\)
 

\begin{array} {|l|c|c|c|}
\hline  & \ce{OCl-} & \ce{HOCl} & \ce{OH–} \\
\hline \text{Initial} & 0.20 & 0 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 0.20-x & x & x \\
\hline \end{array}

 
\[ K_b = \ce{\frac{[HOCl][OH– ]}{[OCl-]}} = \frac{x^2}{(0.20-x)} \]

Assume  `0.20-x~~0.20`  because `x` is negligible:

`3.3 xx 10^(-7)` `= x^2 / (0.20-x)`  
`x` `=sqrt(3.3 xx 10^(−7) xx 0.20)`  
  `= 2.5690 xx 10^{-4}\ text{mol L}^(–1)`  

 
\(\ce{[OH-] = 2.5690 \times 10^{-4} mol L^{-1}}\)

\(\ce{pOH = -log10[OH-] = -log10(2.5690 \times 10^{-4}) = 3.59}\)

\(\therefore \ce{pH = 14-3.59 = 10.41}\)


♦ Mean mark (b) 45%.

CHEMISTRY, M5 2019 HSC 16 MC

At equilibrium, a 1.00 L vessel contains 0.0430 mol of \(\ce{H2}\), 0.0620 mol of \(\ce{I2}\), and 0.358 mole of \(\ce{HI}\). The system is represented by the following equation:

\(\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)}\)

Which of the following is closest to the value of the equilibrium constant, \(K_{eq}\), for this reaction?

  1. 0.0208
  2. 48.1
  3. 134
  4. 269
Show Answers Only

\(B\)

Show Worked Solution
\(K_{eq}\) \(= \dfrac{\ce{[HI]^2}}{\ce{[H_2][I_2]}} \)  
  \(=\dfrac{0.358^2}{(0.0430)(0.0620)}\)  
  \(=48.1\)  

 
\(\Rightarrow B\)

CHEMISTRY, M5 2020 HSC 35

In aqueous solution, iodide ions `(text{I}^-)`react rapidly with iodine `(text{I}_2)` to form triiodide ions `text{I}_(3)^(\ -)`, making the equilibrium system shown in the chemical equation:

`text{I}^(-)(aq)+ text{I}_(2)(aq) ⇌ text{I}_(3)^(\ -)(aq)`

The following relationships can be derived from the reaction mechanism:

`[text{I}^(-)]_(eq)=2[text{I}_(2)]_(eq)`

`[text{I}^(-)]_(i nitial)=4[text{I}_(2)]_(eq)+3[text{I}_(3)^(\ -)]_(eq)`

where 'initial' designates the initial concentration and 'eq' designates the equilibrium concentration.

The absorbance of the solution in the UV-Vis spectrum is given by:

`A=[text{I}_(3)^(\ -)]xx2.76 xx10^(4)`

Determine the value of the equilibrium constant, given that  `A = 0.745`  at equilibrium and  `[text{I}^(-)]_(i nitial )=7.00 xx10^(-4)\ text{mol L}^(-1)`.   (4 marks)

Show Answers Only

`K_text{eq}= 564`

Show Worked Solution
`text{A}` \(\ce{= [I3–]_{eq} \times 2.76 × 10^4}\)
`0.745` \(\ce{= [I3–]_{eq} \times 2.76 × 10^4}\)
\(\ce{[I3–]_{eq}}\) `= 2.70 \times 10^(−5)\ text{mol L}^(–1)`

 

\(\ce{[I– ]_{initial}}\) \(\ce{= 4[I2 ]_{eq} + 3 [I3– ]_{eq}}\)
`7.00 xx 10^(−4)` `= 4 [text{I}_2 ]_text{eq} + (3 xx 2.70 xx 10^(−5))`
\(\ce{[I2 ]_{eq}}\) `=(7.00 xx 10^(−4)-(3 xx 2.70 xx 10^(−5)))/4`
  `= 1.55 xx 10^(−4)\ text{mol L}^(–1)`

 

`[text{I}^– ]_text(eq) = 2 [text{I}_2 ]_text(eq) = 2 xx (1.55 xx 10^(−4)) = 3.10 xx 10^(−4)\ text{mol L}^(–1)`

 

`K_text{eq}` `=[text{I}_3\^(\ -)]_text(eq) /[[text{I}^–]_text(eq) xx [text{I}_2]_text(eq)]`
  `= [2.70 xx 10^(−5)] / [3.10 xx 10^(−4) xx 1.55 xx 10^(−4)]`
  `= 564`

Mean mark 55%.

CHEMISTRY, M5 2022 HSC 13 MC

Nitrosyl bromide decomposes according to the following equation.

\(\ce{2NOBr(g)  \rightleftharpoons  2NO(g) + Br2(g)}\)

A 0.64 mol sample of \(\ce{NOBr}\) is placed in an evacuated 1.00 L flask. After the system comes to equilibrium, the flask contains 0.46 mol \(\ce{NOBr}\).

What are the concentrations of \(\ce{NO}\) and \(\ce{Br2}\) in the flask at equilibrium?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1.2ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1.2ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1.2ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1.2ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline\ce{[NO]}\left( \text{mol L}^{-1}\right) &  \ce{\left[Br_2\right]} \left(\text{mol L}^{-1}\right)\\
\hline \rule{0pt}{2.5ex}0.18 \rule[-1ex]{0pt}{0pt}& 0.09 \\
\hline \rule{0pt}{2.5ex}0.18 \rule[-1ex]{0pt}{0pt}& 0.18 \\
\hline \rule{0pt}{2.5ex}0.36 \rule[-1ex]{0pt}{0pt}& 0.18 \\
\hline \rule{0pt}{2.5ex}0.92 \rule[-1ex]{0pt}{0pt}& 0.46 \\
\hline
\end{array}
\end{align*}

Show Answers Only

`A`

Show Worked Solution

\begin{array} {|l|c|c|c|}
\hline  & \text{NOBr} & \text{NO} & \text{Br}_2 \\
\hline \text{Initial} & 0.64 & 0 & 0 \\
\hline \text{Change} & -0.18 & +0.18 & +0.09 \\
\hline \text{Equilibrium} & 0.46 & 0.18 & 0.09 \\
\hline \end{array}

 
`[text{NO}] = 0.18\ text{mol L}^-1`

`[text{Br}_2] = 0.09\ text{mol L}^-1`

`=> A`

CHEMISTRY, M5 2022 HSC 8 MC

A system is described as follows.

\(\ce{2NaHCO3(s)  \rightleftharpoons Na2CO3(s) + CO2(g) + H2O(g)}\)

What is the equilibrium expression for this system?

  1. `K_{eq}=[\text{CO}_2]`
  2. `K_{eq}=[\text{CO}_2][\text{H}_2\text{O}]`
  3. `K_{eq}=(1)/([\text{CO}_2][\text{H}_2\text{O}])`
  4. `K_{eq}=([\text{Na}_2\text{CO}_3][\text{CO}_2][\text{H}_2\text{O}])/[\text{NaHCO}_3]^(2)`
Show Answers Only

`B`

Show Worked Solution
  • The concentrations of solids and pure liquids are omitted from the equilibrium expression because they have a constant concentration.
  • Thus, the equilibrium expression is:
  •    `K_(eq) = [text{CO}_2] [text{H}_2 text{O}]`

`=> B`

CHEMISTRY, M5 2021 HSC 31

Ammonia is produced according to the following equilibrium equation.

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

There are 4.50 moles of nitrogen gas, 1.00 mole of hydrogen gas and 5.80 moles of ammonia in a 10.0 L vessel. The system is at equilibrium at 298 K. The value of `K_{eq}` at this temperature is 748 .

How many moles of nitrogen gas need to be added to the vessel to increase the amount of ammonia by 0.050 moles?   (4 marks)

Show Answers Only
  `text{N}_2`  `\ \ +\ \ `   `3 text{H}_2`   `⇋`   `\ \ 2 text{NH}_3`
Initial `(4.5 + x)\ text{moles}`   `1.0\ text{moles}`   `5.8\ text{moles}`
Change `− 0.025\ text{moles}`   `− 0.075\ text{moles}`   `+ 0.05\ text{moles}`
Equilibrium   `(4.475 + x)\ text{moles}`   `0.925\ text{moles}`   `5.85\ text{moles}`
Equilibrium 
concentration  		 `[4.475 + x] / 10\ text{mol L}^ (–1)`    `0.0925\ text{mol L}^(–1)`   `0.585\ text{mol L}^(–1)`

 
`K_(eq) = [NH_3]^2 / [[N_2] [H_2]^3] `

`748 = 0.585^2 / [(4.475 + x) / 10 xx 0.0925^3]`

`748 xx (4.475 + x) / 10 xx 0.0925^3 = 0.585^2`

`(4.475 + x)/10` `= 0.585^2 / [748 xx 0.0925^3]`  
`4.475+x` `=[10 xx 0.585^2] / [748 xx 0.0925^3]`  
`x` `=[10 xx 0.585^2] / [748 xx 0.0925^3]-4.475`  
  `=1.3\ text{moles (1 d.p.)}`  

 
∴ 1.3 moles of nitrogen must be added to the equilibrium mixture.

Show Worked Solution
  `text{N}_2`  `\ \ +\ \ `   `3 text{H}_2`   `⇋`   `\ \ 2 text{NH}_3`
Initial `(4.5 + x)\ text{moles}`   `1.0\ text{moles}`   `5.8\ text{moles}`
Change `− 0.025\ text{moles}`   `− 0.075\ text{moles}`   `+ 0.05\ text{moles}`
Equilibrium   `(4.475 + x)\ text{moles}`   `0.925\ text{moles}`   `5.85\ text{moles}`
Equilibrium 
concentration  		 `[4.475 + x] / 10\ text{mol L}^ (–1)`    `0.0925\ text{mol L}^(–1)`   `0.585\ text{mol L}^(–1)`

 
`K_(eq) = [NH_3]^2 / [[N_2] [H_2]^3] `

`748 = 0.585^2 / [(4.475 + x) / 10 xx 0.0925^3]`

`748 xx (4.475 + x) / 10 xx 0.0925^3 = 0.585^2`

`(4.475 + x)/10` `= 0.585^2 / [748 xx 0.0925^3]`  
`4.475+x` `=[10 xx 0.585^2] / [748 xx 0.0925^3]`  
`x` `=[10 xx 0.585^2] / [748 xx 0.0925^3]-4.475`  
  `=1.3\ text{moles (1 d.p.)}`  

 
∴ 1.3 moles of nitrogen must be added to the equilibrium mixture.


♦ Mean mark 44%.