# The value of DH0 for the reaction below is -6535 kJ. __________ kJ of heat are released in the combustion of 16.0 g of C_6H_6(l)? 2C6H6(l) + 15O2 (g) --> 12CO2 (g) + 6H2O(l)

## $2 {C}_{6} {H}_{6} \left(l\right) + 15 {O}_{2} \left(g\right) \to 12 C {O}_{2} \left(g\right) + 6 {H}_{2} O \left(l\right)$ $1.34 \times {10}^{\text{3}}$ $669$ $- 6535$ $5.25 \times {10}^{4}$ $2.68 \times {10}^{3}$

Aug 1, 2017

$\Delta {H}_{\text{rxn}}^{\circ}$ are always quoted per mole of $\text{REACTION}$ as written. And thus $669 \cdot k J$ of energy are $\text{RELEASED}$ from the reaction.

#### Explanation:

We gots.........

$2 {C}_{6} {H}_{6} \left(l\right) + 15 {O}_{2} \left(g\right) \rightarrow 12 C {O}_{2} \left(g\right) + 6 {H}_{2} O \left(l\right) + \Delta$

$\Delta {H}_{\text{rxn}}^{\circ} = - 6535 \cdot k J \cdot m o {l}^{-} 1$, and I reiterate, that this energy value is PER MOLE of reaction as written, i.e. the enthalpy change associated with the combustion of $2 \cdot m o l$ benzene, i.e. a $154 \cdot g$ mass.

We gots, $\frac{16.0 \cdot g}{78.11 \cdot g \cdot m o {l}^{-} 1} = 0.205 \cdot m o l$ with respect to benzene......and thus the heat associated with the combustion of this molar quantity is.........

$\Delta {H}_{\text{rxn}}^{\circ} = 0.205 \cdot m o l \times \frac{1}{2} \times - 6535 \cdot k J \cdot m o {l}^{-} 1 = - 669 \cdot k J$

Note that when we say that this amount of energy is $\text{released from the reaction}$, in effect we are saying that the reaction is exothermic, and $\Delta {H}^{\circ}$ is a $\text{NEGATIVE QUANTITY}$.

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