Question

the enthalpy of formation for $\ce{c6h6(l)}$ is 49.0 kj/mol. consider the reaction.
$\ce{6c(s, graphite) + 3h2(g) -> c6h6(l)}$
is the reaction endothermic or exothermic, and what is the enthalpy of reaction?
use $\delta h_{\text{rxn}} = \sum (\delta h_{f, \text{products}}) - \sum (\delta h_{f, \text{reactants}})$.
\boxed{\text{endothermic; } \delta h_{\text{rxn}} = 49.0 \text{ kj}}
\boxed{\text{endothermic; } \delta h_{\text{rxn}} = -49.0 \text{ kj}}
\boxed{\text{exothermic; } \delta h_{\text{rxn}} = 49.0 \text{ kj}}
\boxed{\text{exothermic; } \delta h_{\text{rxn}} = -49.0 \text{ kj}}

Explanation:

Step1: Identify enthalpies of formation

For reactants: \( \text{C (s, graphite)} \) and \( \text{H}_2\text{(g)} \) have \( \Delta H_f = 0 \, \text{kJ/mol} \) (standard state). For product \( \text{C}_6\text{H}_6\text{(l)} \), \( \Delta H_f = 49.0 \, \text{kJ/mol} \).

Step2: Apply \( \Delta H_{\text{rxn}} \) formula

Use \( \Delta H_{\text{rxn}} = \sum (\Delta H_{f, \text{products}}) - \sum (\Delta H_{f, \text{reactants}}) \).
Products: \( 1 \times 49.0 \, \text{kJ/mol} \).
Reactants: \( 6 \times 0 + 3 \times 0 = 0 \, \text{kJ/mol} \).
Thus, \( \Delta H_{\text{rxn}} = 49.0 - 0 = 49.0 \, \text{kJ} \).

Step3: Determine reaction type

Endothermic reactions have \( \Delta H_{\text{rxn}} > 0 \) (absorb heat), exothermic have \( \Delta H_{\text{rxn}} < 0 \) (release heat). Since \( 49.0 > 0 \), the reaction is endothermic.

Answer:

A. endothermic; \( \Delta H_{\text{rxn}} = 49.0 \, \text{kJ} \)