MolSynapsequestion bank

MolSynapse / Question bank / Thermodynamics and Thermochemistry / Entropy and second law, entropy changes in processes

JEE Main levelSingle correct optionDifficulty 2 of 3td042

An exothermic reaction at constant pressure releases $92.2 kJ$ of heat to a large surroundings at $298 K$ (reversible heat exchange with the surroundings). The entropy change of the surroundings is

A $+ 309.4 J K^{- 1}$
B $- 309.4 J K^{- 1}$
C $+ 0.309 J K^{- 1}$
D $+ 27476 J K^{- 1}$

Answer: A

Δ S_{s u rr} = - Δ H_{sys} / T = 92200/298 = + 309.4 J K^{- 1}

. The surroundings gain heat so their entropy rises. The

0.309

option forgets the kJ to J conversion.

Practice Thermodynamics and Thermochemistry More on Entropy and second law, entropy changes in processes

Previous questionNext question

More questions on Entropy and second law, entropy changes in processes