A) 524.1
B) 41.2
C) \[-262.5\]
D) \[-41.2\]
Correct Answer: B
Solution :
Key Idea: \[\Delta H=\Sigma (\Delta H{{{}^\circ }_{products}})-\Sigma (\Delta H{{{}^\circ }_{reac\tan ts}})\] Given, \[\Delta H_{f}^{o}C{{O}_{2}}(g)=-393.5\text{ }kJ/mol\] \[\Delta H_{f}^{o}=CO(g)=-110.5\text{ }kJ/mol\] \[\Delta H_{f}^{o}{{H}_{2}}O(g)=-241.8\text{ }kJ/mol\] \[C{{O}_{2}}(g)+{{H}_{2}}(g)=-241.8\,kJ/mol\] \[C{{O}_{2}}(g)+{{H}_{2}}(g)\xrightarrow[{}]{{}}CO(g)+{{H}_{2}}O(g)\] \[\Delta H{}^\circ =[(\Delta H_{f}^{o}CO)+(\Delta H_{f}^{o}{{H}_{2}}O)]\] \[-[(\Delta H_{f}^{o}C{{O}_{2}})+(\Delta H_{f}^{o}{{H}_{2}})]\] \[=[(-110.5)+(-241.8)]\] \[-[(-393.5)+(0)]\] \[=-110.5-241.8+393.5\] \[=-352.3+393.5\] \[=+41.2\,kJ\]You need to login to perform this action.
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