When enthalpy and entropy changes for a chemical reaction are 2-5 - 103 cals and 7-4 cals deg-1- respectively- The reaction at 298 K will be -

The correct option is B non spontaneousWe know that : Δ G = Δ H TΔ S Δ G #160;= (2.5× 10^3) (298× 7.4)=+294.8 cals Δ G= + #160;ve ∴ Re ...

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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

When enthalpy...

Question

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

spontaneous

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reversible

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irreversible

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non-spontaneous

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Solution

The correct option is B non-spontaneous
We know that :

$Δ G = Δ H - T Δ S$

$Δ G = (2.5 \times 10^{3}) - (298 \times 7.4) = + 294.8$ cals

$Δ G = + v e$

$∴$ Reaction is non-spontaneous.

Hence, the correct option is D.

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When enthalpy and entropy changes for a chemical reaction are 2.5×103cals and 7.4 cals deg−1, respectively. The reaction at 298 K will be :

The correct option is B non-spontaneousWe know that :ΔG=ΔH−TΔSΔG=(2.5×103)−(298×7.4)=+294.8 calsΔG=+ve∴ Reaction is non-spontaneous.Hence, the correct option is D.

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

The correct option is B non-spontaneous
We know that :

$Δ G = Δ H - T Δ S$

$Δ G = (2.5 \times 10^{3}) - (298 \times 7.4) = + 294.8$ cals

$Δ G = + v e$

$∴$ Reaction is non-spontaneous.

Hence, the correct option is D.