The enthalpy and entropy change for a chemical reaction are -2-5 -103 cal and 7-4 cal K-1 respectively- The reaction at 298 K is-

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. 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

Q. Standard enthalpy and standard entropy change for the oxidation of

{N H}_{3}

298 K

are

- 382.64 k J {m o l}^{- 1}

and

- 145.6 k J {m o l}^{- 1}

respectively. Standard Gibbs energy change for the same reaction at

298 K

is:

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The enthalpy and entropy change for a chemical reaction are −2.5×103cal and 7.4 cal K−1 respectively. The reaction at 298 K is:

The correct option is C spontaneousΔG=ΔH−TΔS ΔG=−2.5×103−298×7.4 ΔG=−4705.2 cal. As the value of ΔG is −4705.2 cal, which is a negative value, so this reaction will be spontaneous. Hence, option A is correct.

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:

The correct option is C spontaneous
$Δ G = Δ H - T Δ S$
$Δ G = - 2.5 \times 10^{3} - 298 \times 7.4$
$Δ G = - 4705.2 c a l$ .
As the value of $Δ G$ is $- 4705.2$ cal, which is a negative value, so this reaction will be spontaneous.
Hence, option A is correct.