SSCE HSC Chemistry Alcohols: When performing industrial reduc

When performing industrial reductions with CO(g), the following equilibrium is of great importance: 2CO(g) ⇌ CO2(g) + C(s) $K_{eq} = 10.00$ at 1095 K A 1.00 L sealed vessel at a temperature of 1095 K contains CO(g) at a concentration of 1.10 x 10^-2 mol L^-1, CO2(g) at a concentration of 1.21 x 10^-3 mol L^-1, and excess solid carbon - HSC - SSCE Chemistry - Question 37 - 2023 - Paper 1

Question 37

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When performing industrial reductions with CO(g), the following equilibrium is of great importance: 2CO(g) ⇌ CO2(g) + C(s) $K_{eq} = 10.00$ at 1095 K A 1.00 L seal... show full transcript

Worked Solution & Example Answer:When performing industrial reductions with CO(g), the following equilibrium is of great importance: 2CO(g) ⇌ CO2(g) + C(s) $K_{eq} = 10.00$ at 1095 K A 1.00 L sealed vessel at a temperature of 1095 K contains CO(g) at a concentration of 1.10 x 10^-2 mol L^-1, CO2(g) at a concentration of 1.21 x 10^-3 mol L^-1, and excess solid carbon - HSC - SSCE Chemistry - Question 37 - 2023 - Paper 1

Step 1

Is the system at equilibrium? Support your answer with calculations.

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Answer

To determine if the system is at equilibrium, we calculate the reaction quotient, Q.

The expression for Q is:

$Q = \frac{[CO_2]}{[CO]^2}$

Substituting the values:

$Q = \frac{1.21 \times 10^{-3}}{(1.10 \times 10^{-2})^2}$

Calculating this, we have:

$Q = \frac{1.21 \times 10^{-3}}{1.21 \times 10^{-4}} = 10.0$

Since $Q = K_{eq}$ (10.0), the system is at equilibrium.

Step 2

Carbon dioxide gas is added to the system above and the mixture comes to equilibrium.

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Answer

At the new equilibrium, the concentrations of CO(g) and CO2(g) are equal.

Let ( x ) be the amount of CO2 added to the system:

Initial concentration of CO2 = ( 1.21 \times 10^{-3} , mol , L^{-1} )
After addition, concentration of CO2 = ( 1.21 \times 10^{-3} + x )
Concentration of CO = ( (1.10 \times 10^{-2}) - 0.0890 )

At equilibrium, we set:

$[CO] = [CO_2] = \frac{1.00}{10.0}$

Thus:

$n(CO_2) = 0.1000 = 0.1 \; (mol \; L^{-1})[CO_2] \qquad (K_{eq} = 10.0)$

Calculating the change in moles:

( n(CO_2) = n(CO_2)_{initial} + x )
Find ( x ):

Since 1 mol of CO2 gives 2 mol of CO:

( n(CO_2) = 1.21 \times 10^{-3} + 0.0890 ) ( n(CO_2) \approx 0.0988 + x )

From this, we find:

( n(CO_2){required} = 0.00445 + \text{amount added} ) ( n(CO_2){added} = 0.00445 + 0.0898 )

In total,

( n(CO_2)_{added} = 0.00445 )

Is the system at equilibrium? Support your answer with calculations.

Carbon dioxide gas is added to the system above and the mixture comes to equilibrium.

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