Question

question 10 (1 point)
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

options: a, b, c, d, e

question 17 (1 point)
use the following information to answer the next question.
nitric oxide (no(g)) is a colourless gas. the main

Explanation:

Step1: Calculate the equilibrium concentrations

• The initial concentrations:
• $[CH_4]_0=\frac{1.5\ mol}{6.0\ L} = 0.25\ M$
• $[H_2O]_0=\frac{1.5\ mol}{6.0\ L}= 0.25\ M$
• At equilibrium, $n(CO) = 0.75\ mol$, so $[CO]=\frac{0.75\ mol}{6.0\ L}=0.125\ M$
• From the stoichiometry of the reaction $CH_4(g)+H_2O(g)

ightleftharpoons CO(g)+3H_2(g)$, if $[CO]=x = 0.125\ M$, then $[H_2]=3x$

• $[H_2]=3\times0.125\ M = 0.375\ M$, and $n(H_2)=0.375\ M\times6.0\ L=2.25\ mol\approx2.3\ mol$
• $[CH_4]=[CH_4]_0 - x=0.25\ M- 0.125\ M = 0.125\ M$
• $[H_2O]=[H_2O]_0 - x=0.25\ M-0.125\ M = 0.125\ M$

Step2: Calculate the equilibrium constant $K_c$

• The equilibrium - constant expression for the reaction $CH_4(g)+H_2O(g)

ightleftharpoons CO(g)+3H_2(g)$ is $K_c=\frac{[CO][H_2]^3}{[CH_4][H_2O]}$

• Substitute the equilibrium concentrations:
• $K_c=\frac{(0.125)(0.375)^3}{(0.125)(0.125)}$
• First, cancel out the $0.125$ terms in the numerator and denominator. Then $K_c=\frac{(0.375)^3}{0.125}$
• $(0.375)^3=0.375\times0.375\times0.375=\frac{3}{8}\times\frac{3}{8}\times\frac{3}{8}=\frac{27}{512}$
• $K_c=\frac{\frac{27}{512}}{\frac{1}{8}}=\frac{27}{512}\times8=\frac{27}{64}\approx0.42$

Answer:

D. 0.42 2.3 mol