The enthalpy and entropy change for a chemical reaction are - 2-5 - 1 0 3 cal and 7-4 cal K -1 respectively- Predict the nature of reaction at 298 K-

Solution: As we know that, ΔG = ΔH T ΔS Given: ΔS = 7.4 cal/K ΔH = 2.5 ×10^3 cal T = 298 K ∴ΔG = 2.5 ×10^3 ( 7.4 × 298 ) ⇒ ...

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Standard XII

Chemistry

Entropy

The enthalpy ...

Question

The enthalpy and entropy change for a chemical reaction are $- 2.5 \times 1 0^{3}$ cal and $7.4$ cal $K^{- 1}$ respectively. Predict the nature of reaction at $298$ K.

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Solution

Solution:-
As we know that,
$Δ G = Δ H - T Δ S$
Given:-
$Δ S = 7.4 c a l / K$
$Δ H = - 2.5 \times 10^{3} c a l$
$T = 298 K$
$∴ Δ G = - 2.5 \times 10^{3} - (7.4 \times 298)$
$\Rightarrow Δ G = - 2500 - 2205.2 = - 4705.2 c a l$
Since the value of $Δ G$ is negative, thus the reaction is spontaneous.

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The enthalpy and entropy change for a chemical reaction are −2.5×103 cal and 7.4 cal K−1 respectively. Predict the nature of reaction at 298 K.

Solution:-As we know that,ΔG=ΔH−TΔSGiven:-ΔS=7.4cal/KΔH=−2.5×103calT=298K∴ΔG=−2.5×103−(7.4×298)⇒ΔG=−2500−2205.2=−4705.2calSince the value of ΔG is negative, thus the reaction is spontaneous.

The enthalpy and entropy change for a chemical reaction are $- 2.5 \times 1 0^{3}$ cal and $7.4$ cal $K^{- 1}$ respectively. Predict the nature of reaction at $298$ K.