ThermodynamicsHard

Question

Given the following entropy values (in J/K-mol) at 298 K and 1 atm H2(g) = 130.6, Cl2(g) = 223.0 and HCl(g) = 186.7. The entropy change (in J/K-mol) for the reaction H2(g) + Cl2(g) → 2HCl(g) is

Options

A.+540.3
B.+727.0
C.−166.9
D.+19.8

Solution

$\Delta_{r}S = \left( 2 \times S_{m,HCl} \right) - \left( S_{m,H_{2}} + S_{m,Cl_{2}} \right)$

$= (2 \times 186.7) - = + 19.8\text{ J/K. mol}$

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