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Multiple Choice
Given the equilibrium reaction 2CO(g) + O2(g) ⇌ 2CO2(g) with Kc = 2.8 × 10^2 at 298 K, and the partial pressures of CO and O2 are 5.168 × 10^-8 torr and 0.988 torr respectively, what is the partial pressure of CO2 in this mixture?
A
5.00 × 10^1 torr
B
3.00 × 10^2 torr
C
2.80 × 10^2 torr
D
1.44 × 10^2 torr
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검증된 단계별 안내
1
Start by writing the expression for the equilibrium constant Kc for the given reaction: \( K_c = \frac{[CO_2]^2}{[CO]^2[O_2]} \).
Convert the given partial pressures from torr to atm, if necessary, since equilibrium constants are typically calculated using concentrations in atm. Use the conversion factor: 1 atm = 760 torr.
Substitute the given partial pressures of CO and O2 into the equilibrium expression. Let \( P_{CO_2} \) be the partial pressure of CO2. The expression becomes: \( K_c = \frac{P_{CO_2}^2}{(5.168 \times 10^{-8})^2 \times 0.988} \).
Rearrange the equation to solve for \( P_{CO_2} \): \( P_{CO_2}^2 = K_c \times (5.168 \times 10^{-8})^2 \times 0.988 \).
Take the square root of both sides to find \( P_{CO_2} \): \( P_{CO_2} = \sqrt{K_c \times (5.168 \times 10^{-8})^2 \times 0.988} \).