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ΔS_sur for H₂+1/2O₂ $\to$ H₂O, ΔH= - 280 KJ at 400 K is

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700 KJ / mol / K

700 J/g / K

0.7 J / mol / K

700 J/mol/K

answer is C.

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Detailed Solution

\begin{array}{l} \Delta {S_{Surr.}} = - \Delta {S_{sys}}\\\\ \Delta {S_{Surr.}}\, = - \left( {\frac{{\Delta {H_{sys}}}}{T}} \right) = - \left( {\frac{{ - 280 \times 1000}}{{400}}} \right)\\\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,700J/mole/K \end{array}

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