A) -44 kcal
B) -88 kcal
C) -22kcal
D) -11 kcal
Correct Answer: C
Solution :
Key Idea: First write the equation for chemical change and then by using following formula enthalpy of formation of \[HCl\] gas can be calculated. \[\Delta {{H}_{f}}=\Sigma \] bond energy of reactants \[-\Sigma \] bond energy of products \[\frac{1}{2}{{H}_{2}}+\frac{1}{2}C{{l}_{2}}\to HCl\,(g)\] Given: Bond energy of \[{{H}_{2}}=104\,kcal\] Bond energy of \[C{{l}_{2}}=58\,kcal\] Bond energy of \[HCl=103\,kcal\] \[\Delta {{H}_{f}}=\frac{1}{2}(104+58)-103\] \[=81-103\] = - 22 kcalYou need to login to perform this action.
You will be redirected in
3 sec