Question

using the table of standard formation enthalpies that youll find under the aleks data tab, calculate the reaction enthalpy of this reaction under standard conditions: 2c₆h₆(l)+15o₂(g)→12co₂(g)+6h₂o(l) round your answer to the nearest kj. kj

Explanation:

Step1: Recall the formula for reaction enthalpy

$$\Delta H_{rxn}=\sum n\Delta H_{f(products)}-\sum n\Delta H_{f(reactants)}$$

Step2: Find the standard - formation enthalpies

From the standard - formation enthalpy table:
$\Delta H_{f}(C_{6}H_{6}(l)) = 49\ kJ/mol$, $\Delta H_{f}(O_{2}(g))=0\ kJ/mol$, $\Delta H_{f}(CO_{2}(g))=- 394\ kJ/mol$, $\Delta H_{f}(H_{2}O(l))=-286\ kJ/mol$

Step3: Calculate the sum of enthalpies of products

For products: $n(CO_{2}) = 12$, $n(H_{2}O)=6$
$\sum n\Delta H_{f(products)}=12\times(-394)+6\times(-286)$
$=-4728-1716=-6444\ kJ$

Step4: Calculate the sum of enthalpies of reactants

For reactants: $n(C_{6}H_{6}) = 2$, $n(O_{2}) = 15$
$\sum n\Delta H_{f(reactants)}=2\times49+15\times0 = 98\ kJ$

Step5: Calculate the reaction enthalpy

$\Delta H_{rxn}=-6444 - 98=-6542\ kJ$

Answer:

$-6542\ kJ$