A) \[-1228.2\text{ }kcal\]
B) \[-1229.3\text{ }kcal\]
C) \[-1232.9\text{ }kcal\]
D) \[-1242.6\text{ }kcal\]
Correct Answer: B
Solution :
\[{{C}_{10}}{{H}_{8}}(s)+12{{O}_{2}}(g)\to 10C{{O}_{2}}(g)+4{{H}_{2}}O(l)\] We know that \[\Delta H=\Delta E+\Delta nRT\] \[\Delta H\]= Heat at constant pressure \[\Delta E\] = Heat at constant volume \[=-1228.2\times {{10}^{3}}cal\] \[\Delta n\]= Change in gaseous moles \[=10-12=-2\] \[T=298K;R=2cal\] \[\therefore \] \[\Delta H=-1228.2\times {{10}^{3}}+(-2)(2)(298)\] \[=-1229392cal\] \[=-1229.392kcal\]
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