smc-3671-20-Calcs given K(eq)

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)

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$4.92\ \text{mol}$

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 18 MC

A reaction mixture, not at equilibrium, is composed of both $\ce{N_2O_4(g)}$ and $\ce{NO_2(g)}$ in a closed container. The reaction quotient for the system, $Q$, is given.

$Q=\dfrac{\left[\ce{NO_2}\right]^2}{\left[\ce{N_2O_4}\right]}$

The rate of the forward reaction is initially greater than the rate of the reverse reaction.

Which diagram shows how $Q$ changes over time for this mixture?

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$B$

Show Worked Solution

→ When calculating the reaction quotient of a chemical reaction, the formula is $\dfrac{products}{reactants}$.

→ The equation for the reaction taking place is:

$\ce{N2O4(g) \rightleftharpoons 2NO2(g)}$

→ As the rate of the forward reaction is greater than the rate of the reverse reaction, $\ce{[NO2]^2}$ will increase and $\ce{[N2O4]}$ will decrease.

→ Hence the value for the reaction quotient, $\dfrac{\left[\ce{NO_2}\right]^2}{\left[\ce{N_2O_4}\right]}$, will increase.

→ As it states both $\ce{N2O4(g)}$ and $\ce{NO2(g)}$ are present in the intial system, the value for $Q$ will not be zero.

$\Rightarrow B$

♦ Mean mark 43%.

CHEMISTRY, M5 2024 HSC 15 MC

The thermal decomposition of lithium peroxide $\ce{(Li_2O_2)}$ is given by the equation shown.

$\ce{2Li_2O_2(s)\rightleftharpoons 2 Li_2 O(s) + O_2(g)}$

Mixtures of $\ce{Li_2O_2}$, $\ce{Li_2O}$ and $\ce{O_2}$ were allowed to reach equilibrium in two identical, closed containers, $\text{P}$ and $\text{Q}$, at the same temperature. The amount of $\ce{Li_2O_2(s)}$ in container $\text{P}$ is double that in container $\text{Q}$ . The amount of $\ce{Li_2O(s)}$ is the same in each container.

What is the ratio of $\left[ \ce{O_2(g)}\right]$ in container $\text{P}$ to $\left[\ce{O_2(g)}\right]$ in container $\text{Q}$?

$1: 1$
$2: 1$
$3: 2$
$5: 4$

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$A$

Show Worked Solution

→ When calculating the $K_{eq}$ of a system, substances in solid states are all given a value of $1$.

→ The equilibrium constant of the above reaction is $K_{eq} = \ce{[O_2(g)]}$.

→ As both mixtures reached equilibrium, the $K_{eq}$ values for each mixture is the same, hence the ratio of $\ce{[O2(g)]}$ in each container is $1:1$.

$\Rightarrow A$

♦♦ Mean mark 39%.

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)} $?

$ \text{0.00 mol L}^{-1}$
$ 4.34 \ × \ 10^{-5}\ \text{mol L}^{-1} $
$6.90 \ × \ 10^{-5}\ \text{mol L}^{-1}$
$8.69 \ × \ 10^{-5}\ \text{mol L}^{-1}$

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$D$

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:

$4.0486 \times 10^{-13}$	$=\dfrac{(x)(2x)^2}{(0.9)^2}$
	$=\dfrac{4x^3}{(0.9)^2}$
$4x^3$	$=3.279 \times 10^{-13}$
$x$	$=4.344 \times 10^{-5}$

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

$\Rightarrow D$

♦♦ Mean mark 34%.

CHEMISTRY, M5 2023 HSC 7 MC

A mixture of 0.8 mol of $ \ce{CO} \text{(g)} $ and 0.8 mol of $ \ce{H2} \text{(g)} $ was placed in a sealed 1.0 L container. The following reaction occurred.

$ \ce{CO} \text{(g)} + 2 \ce{H2} \text{(g)} \rightleftharpoons \ce{CH3} \ce{OH}\text{(g)} $

When equilibrium was established, the mixture contained 0.5 mol of $ \ce{CO} \text{(g)} $.

What amount of $ \ce{H2} \text{(g)} $ was present at equilibrium?

0.2 mol
0.4 mol
0.6 mol
1.0 mol

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$A$

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$ \ce{CO} \text{(g)} + 2 \ce{H2} \text{(g)} \rightleftharpoons \ce{CH3} \ce{OH}\text{(g)} $

\begin{array} {|l|c|c|c|}
\hline & \ce{CO(g)} & \ce{2H2(g)} & \ce{CH3OH} \\
\hline \text{Initial} & 0.8 & 0.8 & 0 \\
\hline \text{Change} & -x & -2x & +x \\
\hline \text{Equilibrium} & 0.5 & 0.2 & 0.03 \\
\hline \end{array}

$x = 0.3$

$\Rightarrow A$

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.

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

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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 10 MC

At a certain temperature, the $\ce{$K_{eq}$}$ for the following reaction is 75.

$ \ce{2O3(g) \rightleftharpoons 3O2(g)}$

0.4 mol of $\ce{O3}$ and 1.2 mol of $\ce{O2}$ were introduced to a 5 L reaction vessel.

Which row of the table correctly identifies the direction of the equilibrium shift and the reason for the shift?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Direction favoured}\rule[-1ex]{0pt}{0pt}& \quad \quad \textit{Reason} \quad \quad \\
\hline
\rule{0pt}{2.5ex}\text{Left}\rule[-1ex]{0pt}{0pt}&Q>K_{e q}\\
\hline
\rule{0pt}{2.5ex}\text{Left}\rule[-1ex]{0pt}{0pt}& Q<K_{e q}\\
\hline
\rule{0pt}{2.5ex}\text{Right}\rule[-1ex]{0pt}{0pt}&Q>K_{e q} \\
\hline
\rule{0pt}{2.5ex}\text{Right}\rule[-1ex]{0pt}{0pt}& Q<K_{e q}\\
\hline
\end{array}
\end{align*}

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$D$

Show Worked Solution

\[\ce{O2: c = \frac{n}{V} = \frac{1.2}{5} = 0.24 mol L^{-1}} \]

\[\ce{O3: c = \frac{n}{V} = \frac{0.4}{5} = 0.08 mol L^{-1}} \]

\[\ce{$K_{eq}$ = \frac{[O2]^{3}}{[O2]^{2}} = 75 (given)}\]

\[\ce{$Q$ = \frac{0.24^{3}}{0.08^{2}} = 2.16} \]

$\ce{Since $Q<K_{eq}$, the reaction will shift right to favour the products until $Q = K_{eq}$} $

$\Rightarrow D$

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, M5 2019 HSC 17 MC

A student makes a solution with a final volume of 200 mL by mixing 100 mL of 0.0500 mol L ¯¹ barium nitrate solution with 100 mL of 0.100 mol L ¯¹ sodium hydroxide solution.

Which row of the table correctly identifies if a precipitate will form under these conditions and the reason?

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`D`

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$\ce{Ba(OH)2(s) \rightleftharpoons Ba^2+(aq) + 2OH-(aq)}$

`K_(sp)=2.55xx10^(-4)`

\[\ce{[Ba^2+] = \frac{(0.05)(0.1)}{(0.2)} = 0.025 M}\]

\[\ce{[OH-] = \frac{(0.1)(0.1)}{(0.2)} = 0.05 M}\]

`Q`	$\ce{= [Ba^2+][OH-]^2}$
	`=(0.025)(0.05)^2`
	`=6.25xx10^(-6)`

Since `Q<K_(sp)` , no precipitate forms.

`=>D`

♦♦♦ Mean mark 36%.

CHEMISTRY, M5 2020 HSC 27

A student makes up a solution of propan-2-amine in water with a concentration of 1.00 mol L ¯¹.

Using structural formulae, complete the equation for the reaction of propan-2-amine with water. (2 marks)
The equilibrium constant for the reaction of propan-2-amine with water is `4.37 xx10^(-4)`.
Calculate the concentration of hydroxide ions in this solution. (3 marks)

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\begin{array} {|l|c|c|c|}
\hline & \ce{C3H7NH2} & \ce{C3H7NH3} & \ce{OH–} \\
\hline \text{Initial} & 1.00 & 0 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 1.00-x & x & x \\
\hline \end{array}

\[ K_b = \ce{\frac{[C3H7NH3+][OH– ]}{[C3H7NH2 ]}} = \frac{x^2}{(1.00-x)} \]

Assume `1.00-x=1.00` because `x` is negligible:

`4.37 xx 10^(−4)`	`= x^2 / 1.00`
`x`	`=sqrt(4.37 xx 10^(−4))`
	`= 0.0209\ text{mol L}^(–1)`

`=> [text{OH}^– ] = 0.0209\ text{mol L}^(–1)`

Show Worked Solution

♦ Mean mark (a) 48%.

\[ K_b = \ce{\frac{[C3H7NH3+][OH– ]}{[C3H7NH2 ]}} = \frac{x^2}{(1.00-x)} \]

Assume `1.00-x=1.00` because `x` is negligible:

`4.37 xx 10^(−4)`	`= x^2 / 1.00`
`x`	`=sqrt(4.37 xx 10^(−4))`
	`= 0.0209\ text{mol L}^(–1)`

`=> [text{OH}^– ] = 0.0209\ text{mol L}^(–1)`

Mean mark (b) 51%.

CHEMISTRY, M6 2020 HSC 14 MC

The equation for the autoionisation of water is shown.

$ \ce{2H2O(l) \rightleftharpoons \ H3O+(aq) + OH-(aq)} $

At 50°C the water ionisation constant, `K_(w)`, is `5.5 xx10^(-14)`.

What is the pH of water at 50°C?

5.50
6.63
6.93
7.00

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`B`

Show Worked Solution

$\ce{$K_w$ = [H3O+][OH–]}$

Since $\ce{[H3O+] = [OH–]}$:

`[text{H}_3 text{O}^+]^2`	`=5.5 xx 10^(−14)`
`[text{H}_3 text{O}^+]`	`= 2.3 xx 10^(−7}\ text{mol L}^(–1)`

`text{pH = −log}_(10) (2.3 xx 10^(−7) ) = 6.63`

`=> 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)

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	`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%.

CHEMISTRY, M5 2021 HSC 19 MC

A quantity of silver nitrate is added to 250.0 mL of 0.100 mol L ¯¹ potassium sulfate at 298 K in order to produce a precipitate. Silver nitrate has a molar mass of 169.9 g mol ¯¹.

What mass of silver nitrate will cause precipitation to start?

0.00510 g
0.186 g
0.465 g
0.854 g

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`C`

Show Worked Solution

The reaction when silver nitrate is added to potassium sulfate is:

`text{2} text{Ag} text{NO}_(3) (aq) + text{K}_(2) text{SO}_(4) (aq) → text{Ag}_(2) text{SO}_(4) (s) + text{2} text{K} text{NO}_(3) (aq)`

Since each `text{K}_(2) text{SO}_(4)` molecule has 1 sulfate ion

`[text{SO}_(4) ^(2-)] = text{K}_(2) text{SO}_(4) = 0.100\ text{mol L}^(-1)`

`text{Ag}_(2) text{SO}_(4) (s) ⇋ 2text{Ag}^(+) (aq) + text{SO}_(4) ^(\ 2-) (aq)`

`text{K}_(sp) = [text{Ag}^(+)]^2 [text{SO}_(4) ^(2-)]`

From the data sheet:

`text{K}_(sp) = text{1.20 x 10}^-5`

`text{1.20 x 10}^-5 = [text{Ag}^(+)]^2 xx [text{SO}_(4) ^(\ 2-)]`

`text{1.20 x 10}^-5 = [text{Ag}^(+)]^2 xx [text{0.100}]`

`[text{Ag}^(+)] = 0.01095…\ text{mol L}^(-1)`

`[text{Ag}^(+)] = [text{AgNO}_(3)]`

`text{n(AgNO}_(3)) = text{c} xx text{V} = 0.01095… xx 0.250 = 0.00273…\ text{mol}`

`text{m(AgNO}_(3)) = text{n} xx text{MM} = 0.00273… xx 169.9 = 0.465\ text{g}`

`=> C`

♦ Mean mark 45%.

\(K_{eq}\)	\(=\dfrac{\ce{[H2O(g)][CO(g)]}}{\ce{[H2(g)][CO2(g)]}}\)
\(1.600\)	\(=\dfrac{0.2 \times (1.8+x)}{1.2 \times 0.7}\)
\(1.8 + x\)	\(=1.6 \times \dfrac{1.2 \times 0.7}{0.2}\)
\(x\)	\(=6.72-1.8\)
	\(=4.92\ \text{mol}\)

\(4.0486 \times 10^{-13}\)	\(=\dfrac{(x)(2x)^2}{(0.9)^2}\)
	\(=\dfrac{4x^3}{(0.9)^2}\)
\(4x^3\)	\(=3.279 \times 10^{-13}\)
\(x\)	\(=4.344 \times 10^{-5}\)

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

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