Question

calculate the enthalpy, in kj/mol, for the combustion of the hydrocarbon from the data from the previous steps.
q_cal = 6191.78 j n = 0.0019243682 mol
δh = ? kj/mol
enter either a + or - sign and the magnitude. give your answer to three significant figures.

Explanation:

Step1: Recall the relationship for enthalpy of combustion

For a combustion reaction, the heat absorbed by the calorimeter ($q_{cal}$) is equal in magnitude but opposite in sign to the heat released by the reaction ($q_{rxn}$). So, $q_{rxn} = -q_{cal}$. Then, the enthalpy change $\Delta H$ is given by $\Delta H=\frac{q_{rxn}}{n}$, where $n$ is the moles of the hydrocarbon.
First, convert $q_{cal}$ from joules to kilojoules: $q_{cal}=6191.78\ J = 6191.78\times10^{- 3}\ kJ=6.19178\ kJ$.
Then, $q_{rxn}=-q_{cal}=-6.19178\ kJ$.

Step2: Calculate $\Delta H$

Now, use the formula $\Delta H=\frac{q_{rxn}}{n}$. Substitute $q_{rxn}=-6.19178\ kJ$ and $n = 0.0019243682\ mol$ into the formula:
$\Delta H=\frac{-6.19178\ kJ}{0.0019243682\ mol}$
Calculate the value: $\frac{-6.19178}{0.0019243682}\approx - 3217.5\ kJ/mol$. Rounding to three significant figures, we get $\Delta H\approx - 3.22\times10^{3}\ kJ/mol$ or $-3220\ kJ/mol$ (but in three significant figures, it is $-3.22\times10^{3}$ which can be written as $-3220$ when considering significant figures, but more accurately, $-3.22\times10^{3}$ is $-3220$ with three significant figures? Wait, let's recalculate the division more accurately.

Wait, $6.19178\div0.0019243682$: Let's do the division. $6.19178\div0.0019243682\approx3217.5$. So with three significant figures, that's $3220$ (since the fourth digit is 7, which rounds up the third digit 1 to 2? Wait, no: 3217.5, the first three significant figures are 3,2,1 and the next digit is 7, so we round the third digit up: 3220. So $\Delta H=-3220\ kJ/mol$ (or in scientific notation, $-3.22\times10^{3}\ kJ/mol$)

Answer:

$-3220$ (or $\boldsymbol{-3.22\times10^{3}}$)