using the information in the table to the right, calculate the enthalpy of combustion of 1 mole of acetylene…

Explanation:

Step1: Recall the formula for enthalpy of reaction

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

Step2: Identify $\Delta H_{f}$ values and coefficients

For products:

• $n_{CO_2} = 4$, $\Delta H_{f}(CO_2)=- 393.5$ kJ/mol
• $n_{H_2O}=2$, $\Delta H_{f}(H_2O)= - 241.8$ kJ/mol For reactants:
• $n_{C_2H_2}=2$, $\Delta H_{f}(C_2H_2)=226.8$ kJ/mol
• $\Delta H_{f}(O_2) = 0$ (standard - state element)

Step3: Calculate $\sum n\Delta H_{f}(products)$

$\sum n\Delta H_{f}(products)=4\times(-393.5)+2\times(-241.8)$ $=-1574 - 483.6=-2057.6$ kJ/mol

Step4: Calculate $\sum n\Delta H_{f}(reactants)$

$\sum n\Delta H_{f}(reactants)=2\times226.8+5\times0 = 453.6$ kJ/mol

Step5: Calculate $\Delta H_{rxn}$

$\Delta H_{rxn}=-2057.6 - 453.6=-2511.2$ kJ/mol But we want the enthalpy of combustion of 1 mole of acetylene. Since the coefficient of $C_2H_2$ in the reaction is 2, we divide $\Delta H_{rxn}$ by 2. $\Delta H=\frac{-2511.2}{2}=-1255.6$ kJ/mol

Answer:

$-1255.6$ kJ/mol