Question

question 16 (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
a. | 1.22 | 2.5 ml
b. | 1.79 | 3.5 ml
c | 0.25 | 0 ml
d. | 0.42 | 2.3 mol
e. | 0.50 | 1.5 mol

a
b
c
d
e

Explanation:

Step1: Calculate the equilibrium concentrations

• The reaction is \(CH_{4}(g)+H_{2}O(g)

ightleftharpoons CO(g) + 3H_{2}(g)\)

• Initial moles of \(CH_{4}=1.5\ mol\), initial moles of \(H_{2}O = 1.5\ mol\), volume \(V = 6.0\ L\)
• At equilibrium, moles of \(CO=n_{CO}=0.75\ mol\)
• From the stoichiometry of the reaction, if \(x\) moles of \(CO\) are formed, \(x\) moles of \(CH_{4}\) and \(x\) moles of \(H_{2}O\) react. Here \(x = 0.75\ mol\)
• Moles of \(CH_{4}\) at equilibrium: \(n_{CH_{4}}=(1.5 - 0.75)\ mol=0.75\ mol\)
• Moles of \(H_{2}O\) at equilibrium: \(n_{H_{2}O}=(1.5 - 0.75)\ mol = 0.75\ mol\)
• Moles of \(H_{2}\) at equilibrium: \(n_{H_{2}}=3\times0.75\ mol = 2.25\ mol\approx2.3\ mol\) (rounding to two significant figures)
• Concentrations: \(C=\frac{n}{V}\)
• \([CH_{4}]=\frac{0.75\ mol}{6.0\ L}=0.125\ M\), \([H_{2}O]=\frac{0.75\ mol}{6.0\ L}=0.125\ M\), \([CO]=\frac{0.75\ mol}{6.0\ L}=0.125\ M\), \([H_{2}]=\frac{2.25\ mol}{6.0\ L}=0.375\ M\)

Step2: Calculate the equilibrium constant \(K_{c}\)

• The formula for \(K_{c}\) is \(K_{c}=\frac{[CO][H_{2}]^{3}}{[CH_{4}][H_{2}O]}\)
• Substitute the values: \(K_{c}=\frac{(0.125)\times(0.375)^{3}}{(0.125)\times(0.125)}\)
• First, simplify \(\frac{(0.125)\times(0.375)^{3}}{(0.125)\times(0.125)}=\frac{(0.375)^{3}}{0.125}\)
• \((0.375)^{3}=0.375\times0.375\times0.375 = 0.052734375\)
• \(\frac{0.052734375}{0.125}=0.421875\approx0.42\)

Answer:

D. \(0.42\), \(2.3\ mol\)