When the heat of reaction at constant pressure is -2-5- 103 cal and entropy change is 7-4 cal deg-1 at 25-C- the reaction is predicted as-

The enthalpy and entropy change for a chemical reaction are $- 2.5 \times 10^{3}$ cal and 7.4 cal $d e g^{- 1}$ respectively. The reaction at 298 K will be:

Q. For a reaction at

25^{\circ} C

enthalpy change

(Δ H)

and entropy change

(Δ S)

are

- 11.7 \times 10^{3} {J mol}^{- 1}

and

- 105 {J mol}^{- 1} K^{- 1}

respectively. The reaction is:

Q. The enthalpy and entropy change for a chemical reaction are

- 2.5 \times 10^{3} c a l

and

7.4 c a l K^{- 1}

respectively. The reaction at

298 K

is:

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When the heat of reaction at constant pressure is −2.5×103 cal and entropy change is 7.4 cal deg−1 at 25∘C, the reaction is predicted as:

The correct option is C spontaneousFor the reaction at constant pressure,△H=−2.5×103cal△S=7.4 cal deg−1∴△G=△H−T△S=−2.5×103−298×7.4ΔG=−2.5×103−2205.2△G=−4705.2 cal∴ Since ΔG is negative so the reaction is spontaneous.

When the heat of reaction at constant pressure is $- 2.5 \times 10^{3}$ cal and entropy change is $7.4 c a l d e g^{- 1}$ at $25^{\circ} C$ , the reaction is predicted as: