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?

One "mole" of NH_(4)HS(s) was allowed to decompose in a 1-L container

1.	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?
Answer» `1//4` `3//4` `1//8` `4` Solution :`{:(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)`