Consider this chemical reaction: $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ $H_2(g) + I_2(g) rightleftharpoons 2HI(g)$ . At equilibrium in a particular experiment, the concentrations of H2, I2, and HI were 0.15 M, 0.033 M, and 0.55 M, respectively. What is the value of $K_{e q}$ $K_(eq)$ for this reaction?

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Consider this chemical reaction: $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ $H_2(g) + I_2(g) rightleftharpoons 2HI(g)$ . At equilibrium in a particular experiment, the concentrations of H2, I2, and HI were 0.15 M, 0.033 M, and 0.55 M, respectively. What is the value of $K_{e q}$ $K_(eq)$ for this reaction?

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Answer 1 · 2016-12-01T10:28:06

$K_{eq} > 50$ $K_"eq">50$

Explanation:

For the reaction:

$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ $H_2(g) + I_2(g) rightleftharpoons2HI(g)$

$K_{eq} = \frac{{[H I (g)]}^{2}}{[H_{2} (g)] [I_{2} (g)]}$ $K_"eq"=[HI(g)]^2/([H_2(g)][I_2(g)])$

$=$ $=$ $\frac{{(0.55 \cdot m o l \cdot L^{- 1})}^{2}}{0.15 \cdot m o l \cdot L^{- 1} \times 0.033 \cdot m o l \cdot L^{- 1}}$ $(0.55*mol*L^-1)^2/(0.15*mol*L^-1xx0.033*mol*L^-1)$

$=$ $=$ $? ?$ $??$

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Consider this chemical reaction: H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) rightleftharpoons 2HI(g). At equilibrium in a particular experiment, the concentrations of H2, I2, and HI were 0.15 M, 0.033 M, and 0.55 M, respectively. What is the value of KeqK_(eq) for this reaction?

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