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Multiple Choice
At 445 ºC calculate the equilibrium concentration of HI when 0.500 M HI, 0.0200 M H2, and 0.0200 M I2 are placed in a sealed flask and allowed to reach equilibrium. Kc for this reaction is 50 at 445 ºC. H2 (g) + I2 (g) ⇆ 2 HI (g)
A
0.020 M
B
0.060 M
C
0.460 M
D
0.421 M
E
0.579 M
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1
Write the balanced chemical equation for the reaction: \( \text{H}_2 (g) + \text{I}_2 (g) \rightleftharpoons 2 \text{HI} (g) \).
Set up the expression for the equilibrium constant \( K_c \) for the reaction: \( K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \).
Introduce a change variable \( x \) to represent the change in concentration of \( \text{H}_2 \) and \( \text{I}_2 \) as the system reaches equilibrium. The change for \( \text{HI} \) will be \( +2x \), and for \( \text{H}_2 \) and \( \text{I}_2 \) it will be \( -x \).
Substitute the equilibrium concentrations into the \( K_c \) expression: \( K_c = \frac{(0.500 + 2x)^2}{(0.0200 - x)(0.0200 - x)} \) and solve for \( x \) using the given \( K_c = 50 \).