## Related questions with answers

Consider the following reaction at equilibrium:

$2 \mathrm{H}_2(g)+\mathrm{S}_2(g) \rightleftarrows 2 \mathrm{H}_2 \mathrm{~S}(g)+\text { heat }$

In a 10.0-L container, an equilibrium mixture contains $2.02 \mathrm{~g}$ of $\mathrm{H}_2, 10.3 \mathrm{~g}$ of $\mathrm{S}_2$, and $68.2 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{~S}$. (7.1, 7.2, 13.2, 13.3, $13.4,13.5)$ a. What is the numerical value of $K_{\mathrm{c}}$ for this equilibrium mixture? b. If more $\mathrm{H}_2$ is added to the equilibrium mixture, how will the equilibrium shift? c. How will the equilibrium shift if the mixture is placed in a 5.00-L container with no change in temperature? d. If a 5.00-L container has an equilibrium mixture of $0.300 \mathrm{~mol}$ of $\mathrm{H}_2$ and $2.50 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{~S}$, what is the $\left[\mathrm{S}_2\right]$ if temperature remains constant?

Solution

VerifiedGiven is the balanced reaction at equilibrium:

$\ce{2\ H_2(g) + S_2(g) <=> 2\ H_2S(g) + heat}$

For this equilibrium mixture given is the following data from which we have to calculate the asked:

$V= 10.0 \text{ L}$

$m(\ce{H_2})= 2.02 \text{ g}$

$m(\ce{S_2})= 10.3 \text{ g}$

$m(\ce{H_2S})= 68.2 \text{ g}$

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