Question

calculate the enthalpy of combustion of n₂h₄ using the data from the previous steps.
0.03119 mol n₂h₄ qᵣₓₙ = −20,744.4 j
δh₍comb₎ = ? (kj/mol)
enter either a + or - sign and the magnitude. use significant figures.

Explanation:

Step1: Recall the formula for enthalpy of combustion

The enthalpy of combustion ($\Delta H_{\text{comb}}$) is calculated by dividing the heat of the reaction ($q_{\text{rxn}}$) by the moles of the substance combusted ($n$). The formula is $\Delta H_{\text{comb}}=\frac{q_{\text{rxn}}}{n}$. Also, we need to convert the units from joules to kilojoules (since $1\space kJ = 1000\space J$).

Step2: Substitute the given values

We are given $q_{\text{rxn}}=- 20744.4\space J$ and $n = 0.03119\space mol$. First, convert $q_{\text{rxn}}$ to kilojoules: $q_{\text{rxn}}=-20744.4\space J\times\frac{1\space kJ}{1000\space J}=- 20.7444\space kJ$.

Then, use the formula $\Delta H_{\text{comb}}=\frac{q_{\text{rxn}}}{n}=\frac{- 20.7444\space kJ}{0.03119\space mol}$.

Step3: Calculate the value

$\frac{-20.7444}{0.03119}\approx - 665.1\space\frac{kJ}{mol}$ (we consider significant figures: the given values have 5 significant figures for moles and 5 for $q_{\text{rxn}}$, so the result should be reported with appropriate significant figures. The calculation gives approximately - 665.1 when rounded to a reasonable number of significant figures).

Answer:

-665.1