Consider the following reaction at equilibrium:

2H2(g)+S2(g)2H2 S(g)+ heat 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 g2.02 \mathrm{~g} of H2,10.3 g\mathrm{H}_2, 10.3 \mathrm{~g} of S2\mathrm{S}_2, and 68.2 g68.2 \mathrm{~g} of H2 S\mathrm{H}_2 \mathrm{~S}. (7.1, 7.2, 13.2, 13.3, 13.4,13.5)13.4,13.5) a. What is the numerical value of KcK_{\mathrm{c}} for this equilibrium mixture? b. If more H2\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 mol0.300 \mathrm{~mol} of H2\mathrm{H}_2 and 2.50 mol2.50 \mathrm{~mol} of H2 S\mathrm{H}_2 \mathrm{~S}, what is the [S2]\left[\mathrm{S}_2\right] if temperature remains constant?


Answered 1 year ago
Answered 1 year ago
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Given is the balanced reaction at equilibrium:

2 HX2(g)+SX2(g)2 HX2S(g)+heat\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 LV= 10.0 \text{ L}

m(HX2)=2.02 gm(\ce{H_2})= 2.02 \text{ g}

m(SX2)=10.3 gm(\ce{S_2})= 10.3 \text{ g}

m(HX2S)=68.2 gm(\ce{H_2S})= 68.2 \text{ g}

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