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 products and their $\Delta H_f$ values

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

Step3: Identify reactants and their $\Delta H_f$ values

For reactants:

• $n_{C_2H_2}=2$, $\Delta H_f(C_2H_2)=226.8$ kJ/mol
• $n_{O_2}=5$, $\Delta H_f(O_2) = 0$ (standard - state element) $\sum n\Delta H_f(reactants)=2\times226.8+5\times0 = 453.6$ kJ/mol

Step4: Calculate $\Delta H_{rxn}$

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

Answer:

$-1255.6$ kJ/mol