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

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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