InterviewSolution
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InterviewSolution
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One "mole" of `NH_(4)HS(s)` was allowed to decompose in a `1-L` container at `200^(@)C`. It decomposes reversibly to `NH_(3)(g)` and `H_(2)S(g). NH_(3)(g)` further undergoes decomposition to form `N_(2)(g)` and `H_(2)(g)`. Finally, when equilibrium was set up, the ratio between the number of moles of `NH_(3)(g)` and `H_(2)(g)` was found to be `3`. `NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g), K_(c)=8.91xx10^(-2) M^(2)` `2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g), K_(c)=3xx10^(-4) M^(2)` Answer the following: What is the "mole" fraction of hydrogen gas in the equilibrium mixture in the gas phase?A. `1//4`B. `3//4`C. `1//8`D. `4` |
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Answer» Correct Answer - B `{:(NH_(4)HS(s),hArr,NH_(3)(g),+,H_(2)S(g),),(,,27xx10^(-2),,X):}` `K_(c)=27xx10^(-2)X=8.91xx10^(-2)` `X=(8.91xx10^(-2))/(2.7xx10^(-1))=3.3xx10^(-1)` Number of mol of `NH_(3)=0.27` `H_(2)S=0.33` `N_(2)=0.03` `H_(2)=0.09` `X_(H_(2))=0.09/0.72=1/8` `{:(2NH_(3)(g),hArr,N_(2)(g),+,3H_(2)(g)),(3a,,1/3a,,a):}` `K_(c)=((a//3)a^(3))/((3a)^(2))=a^(4)/3xx1/(9a^(2))` `a^(2)/27=3xx10^(-4)` `:. a^(2)=81xx10^(-4)` `a=9xx10^(-2)` |
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