A) - 372 kcal
B) 162 kcal
C) -240 kcal
D) 183.5 kcal
Correct Answer: A
Solution :
Key Idea 2 Use the following formula to find heat of combustion of ethane. \[\Delta {{H}^{o}}=(\Sigma H_{{{f}_{products}}}^{o})-\,(\Sigma \,\,H_{{{f}_{reactants}}}^{o})\] Ethane undergoes combustion according to following reaction \[{{C}_{2}}{{H}_{6}}(g)+\frac{7}{2}{{O}_{2}}(g)\to 2C{{O}_{2}}(g)+3{{H}_{2}}O(g)\] Given, heat of formation of ethane = -21.1 kcal Heat of formation of \[C{{O}_{2}}=-94.1\,kcal\] Heat of formation of \[{{H}_{2}}O=-68.3\,kcal\] Heat of formation of \[{{O}_{2}}=0\] \[\therefore \] \[{{H}^{o}}=[2H_{{{f}_{(C{{O}_{2}})}}}^{o}+3H_{{{f}_{({{H}_{2}}O)}}}^{o}]\] \[-[H_{{{f}_{\,({{C}_{2}}{{H}_{6}})}}}^{o}+\frac{7}{2}H_{{{f}_{(C{{O}_{2}})}}}^{o}]\] \[=[2\times (-94.1)+3\times (-68.3)]\] \[-[-2.11+(7/2\times 0)]\] \[=(-188.2-204.9)-(-21.1)\] \[=-393.1+21.1\] = - 372 kcal
You need to login to perform this action.
You will be redirected in
3 sec
