Given the following entropy values (in J/K-mol) at 298 K and 1 atm (g) = 130.6, (g) = 223.0 and (g) = 186.7. The entropy change (in J/K-mol) for the reaction: (g) + (g) - 2(g) is

Question

Given the following entropy values (in J/K-mol) at 298 K and 1 atm $H_2$(g) = 130.6, $Cl_2$(g) = 223.0 and $HCl$(g) = 186.7. The entropy change (in J/K-mol) for the reaction: $H_2$(g) + $Cl_2$(g) -> 2$HCl$(g) is

Prashant Kotian · Accepted Answer

Correct answer: D. Δ S = 2 * S() - [S() + S()] = 2 * 186.7 - [130.6 + 223.0] = 373.4 - 353.6 = +19.8 J/K-mol.

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