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At 320 K, a gas A(2) is 20% dissociated ...

At 320 K, a gas `A_(2)` is 20% dissociated to A(g). The standard free energy change at 320 K and 1 atm in `J mol^(-1)` is approximately : `(R =8.314 Jk^(-1)"mol"^(-1), "In" 2=0.693, "In" 3 = 1.098)`

A

4763

B

2068

C

1844

D

4281

Text Solution

Verified by Experts

The correct Answer is:
A
`A_(2)(g) rarr 2A(g)`
`P_(A_(2)) = 0.8/1.2 P_(total) P_(A) = 0.4/0.1 P_(total)`
`K_(p) = p^(2)_(A)/p_(A_(2)) implies DeltaG^(@) = -RT lnK_(p)`
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