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 × 103 cal respectively?

Question

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 × 103 cal respectively? (A) Reversible (B) Spontaneous (C) Non-Spontaneous (D) Irreversible

Tardigrade Admin · Accepted Answer

Given, T=298 K Entropy change (Δ S)=7.4 cal K -1 Enthalpy change (Δ H)=-2.5 × 103 cal As we know that, Δ G=Δ H-T Δ S ∴ Δ G=-2.5 × 103-(298 × 7.4) arrow Δ G=-2.5 × 103-2.2 × 103 cal Δ G=-4.7 × 103 cal Since, the value of Δ G is negative and thus, the reaction is spontaneous in nature.

Q. What is the nature of reaction at 298K, if the entropy change and enthalpy change for a chemical reaction are 7.4calK−1 and −2.5×103cal respectively?

Reversible

Spontaneous

Non-Spontaneous

Irreversible

Given, T=298K Entropy change (ΔS)=7.4 cal K−1 Enthalpy change (ΔH)=−2.5×103 cal As we know that, ΔG=ΔH−TΔS ∴ΔG=−2.5×103−(298×7.4) →ΔG=−2.5×103−2.2×103cal ΔG=−4.7×103cal Since, the value of ΔG is negative and thus, the reaction is spontaneous in nature.

Q. What is the nature of reaction at $298 K$ , if the entropy change and enthalpy change for a chemical reaction are $7.4 c a l K^{- 1}$ and $- 2.5 \times 1 0^{3} c a l$ respectively?