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Question

The equilibrium constant for the reaction is 9.40 at $$900^{\circ}C$$. $$S_{2}(g)+C(s)\rightleftharpoons CS_{2}(g).$$
The pressure of two gases at equilibrium, when 1.42 atm of $$ S_{2} $$ and excess of $$C(s)$$ come to equilibrium is :


A
PCS2=1.284atm, PS2=0.1365atm
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B
PCS2=0.1365atm, PS2=1.284atm
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C
PCS2=1.345atm, PS2=0.1456atm
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D
PCS2=1.444atm, PS2=0.1857atm
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Solution

The correct option is A $$P_{CS_2} = 1.284 atm$$, $$P_{S_2}=0.1365 atm$$
Initial pressure of $$S_2$$ is 1.42 atm. The equilibrium pressures of $$S_2$$ and $$CS_2$$ are 1.42-x and x respectively.
The expression for the equilibrium constant is $$K=\frac {P_{CS_2}}{P_{S_2}}=\frac {x}{1.42-x}=9.4$$
Thus $$x=1.284 atm$$ and $$1.42-x=0.1365atm$$.
Thus $$P_{CS_2} = 1.284 atm$$

Chemistry

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