Question

use the following information to answer the next question.
carbon monoxide (co(g)) is a colourless and odourless gas. it is extremely toxic but has wide applications in chemical manufacturing. it is produced in the following equilibrium reaction:
ch₄(g) + h₂o(g) ⇌ co(g) + 3h₂(g)

at 450 k, the initial concentration of ch₄(g) is 1.5 mol and that of h₂o(g) is 1.5 mol. the reaction takes place in a rigid 6.0 l container. at equilibrium, the amount of co(g) present is 0.75 mol. the equilibrium constant is i and the amount of h₂(g) present at equilibrium is ii. the above statement is completed by the information in row:

row	i	ii
b.	1.79	3.5 ml
c	0.25	0 ml
d.	0.42	2.3 mol
e.	0.50	1.5 mol

Explanation:

Step1: Calculate initial concentrations

Initial concentration of \( \text{CH}_4 \): \( \frac{1.5\ \text{mol}}{6.0\ \text{L}} = 0.25\ \text{mol/L} \)
Initial concentration of \( \text{H}_2\text{O} \): \( \frac{1.5\ \text{mol}}{6.0\ \text{L}} = 0.25\ \text{mol/L} \)

Step2: Set up ICE table

Species	Initial (mol/L)	Change (mol/L)	Equilibrium (mol/L)
\( \text{H}_2\text{O} \)	0.25	\( -x \)	\( 0.25 - x \)
\( \text{CO} \)	0	\( +x \)	\( x \)
\( \text{H}_2 \)	0	\( +3x \)	\( 3x \)

Equilibrium amount of \( \text{CO} \): \( 0.75\ \text{mol} \), so equilibrium concentration of \( \text{CO} \): \( \frac{0.75\ \text{mol}}{6.0\ \text{L}} = 0.125\ \text{mol/L} \). Thus, \( x = 0.125\ \text{mol/L} \).

Step3: Calculate equilibrium concentrations

• \( [\text{CH}_4]_{\text{eq}} = 0.25 - 0.125 = 0.125\ \text{mol/L} \)
• \( [\text{H}_2\text{O}]_{\text{eq}} = 0.25 - 0.125 = 0.125\ \text{mol/L} \)
• \( [\text{CO}]_{\text{eq}} = 0.125\ \text{mol/L} \)
• \( [\text{H}_2]_{\text{eq}} = 3x = 3(0.125) = 0.375\ \text{mol/L} \)

Step4: Calculate equilibrium constant \( K_c \)

\( K_c = \frac{[\text{CO}][\text{H}_2]^3}{[\text{CH}_4][\text{H}_2\text{O}]} \)
Substitute values:
\( K_c = \frac{(0.125)(0.375)^3}{(0.125)(0.125)} \)
Simplify: \( K_c = \frac{(0.375)^3}{0.125} = \frac{0.052734375}{0.125} \approx 0.42 \)

Step5: Calculate equilibrium amount of \( \text{H}_2 \)

Concentration of \( \text{H}_2 \) at equilibrium: \( 0.375\ \text{mol/L} \)
Amount: \( 0.375\ \text{mol/L} \times 6.0\ \text{L} = 2.25\ \text{mol} \approx 2.3\ \text{mol} \)

Answer:

D. 0.42, 2.3 mol