What is the nature of reaction at 298 K, if the entropy change and enthalpy change for a chemical reaction are 7 . 4 cal `K^(-1) and - 2 . 5 xx 10 ^(3) ` cal respectively ?

Options

A. Reversible
B. Spontaneous
C. Non-spontaneous
D. Irreversible

Answer

B

Explanation

Given,
T = 298 K
Entropy change `(Delta S) = 7 . 4 cal K^(-1)`
Enthalpy change `(Delta H) = - 2.5 xx 10^(3) cal`
As we know that,
`Delta G =Delta H - T Delta S`
`:. Delta G = - 2 . 5 xx 10^(3) - (298 xx 7 . 4)`
`rArr Delta = - 2 . 5 xx 10^(3) - 2.2 xx 10 ^(3) cal`
`Delta G = - 4 . 7 xx 106(3) cal`
Since, the value of `Delta G` is negative and thus, the reaction is spontaneous in nature.