InterviewSolution
Saved Bookmarks
| 1. |
The combustion of `1 mol` of benzene takes place at `298 K` and `1 atm`. After combustion, `CO_(2)(g)` and `H_(2)O(l)` are produced and `3267.0 kJ` of heat is librated. Calculate the standard entalpy of formation, `Delta_(f)H^(Θ)` of benzene Given: `Delta_(f)H^(Θ)CO_(2)(g) = -393.5 kJ mol^(-1)` `Delta_(f)H^(Θ)H_(2)O(l) = -285.83 kJ mol^(-1)`.A. `-48.51` kJ/MoleB. `48.51` kJ/MoleC. `-24.5 kJ`D. `24.5 kJ` |
|
Answer» Correct Answer - B `DeltaH=H_(p)-H_(R)` `6C+3H_(2) rarr C_(6)H_(6)" "DeltaH= ?` |
|