1. Tardigrade
  2. Question
  3. Chemistry
  4. When enthalpy and entropy change for a chemical reaction are -2.5 × 103 cal and 7.4 cal deg -1 respectively. Predict that reaction at 298 K is

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Q. When enthalpy and entropy change for a chemical reaction are $-2.5 \times 10^{3}$ cal and $7.4\, cal \,deg ^{-1}$ respectively. Predict that reaction at $298\, K$ is

Thermodynamics

Solution:

$\Delta G=\Delta H-T \Delta S$

Given $\Delta H=-$ ve and $\Delta S=+$ ve. So $\Delta G$ must be negative, hence reaction will be spontaneous.