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For the reaction A(g)+2B(g)hArr2D(g),Del...

For the reaction `A(g)+2B(g)hArr2D(g),DeltaG^(0)=2.kJ*mol^(-1)` at 500K. What is the value of `K_(p)` for the reaction `(1)/(2)A(g)+B(g)hArr D(g)` at that temperature?

Text Solution

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From the equilibrium `DeltaG^(0)=-RTlnK_(p)`, we get, `lnK_(p)=-(DeltaG^(0))/(RT)=-(2000J*mol^(-1))/(8.314J*K^(-1)*mol^(-1)xx500K)=-0.4811`
`therefore K_(p)=0.6181`
`therefore`For the given reaction, `K_(p)=((p_(D))^(2))/(p_(A)xx(p_(B))^(2))`
If the equilibrium constant for the reaction
`(1)/(2)A(g)+B(g)hArrD(g)` be `K_(p)'`, then `K_(p)'=(p_(D))/(p_(A)^(1//2)xxp_(B))`
`therefore K_(p)'=sqrt(K_(p))" "thereforeK_(p)'=sqrt(0.6181)=0.7862`
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