# The combustion of propane (C_3H_8) produces 248 of energy per mole of propane burned. How much heat energy will be released when 1 000 L of propane are burned at STP?

Apr 2, 2017

Approx......${10}^{5} \cdot k J$ are evolved...........

#### Explanation:

We can represent propane combustion by the following equation:

${C}_{3} {H}_{8} \left(g\right) + {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(l\right) + 2220 \cdot k J$

I took the heat of combustion values from this site. Your question does not even quote UNITS..........

And, yes, I know that we would normally write $\Delta {H}_{\text{combustion}}^{\circ} = - 2220 \cdot k J \cdot m o l .$ We KNOW that hydrocarbon combustion is exothermic, so we can represent energy, $\Delta$, as a reaction PRODUCT, for the purposes of answering this question.

And we have a $1000 \cdot L$ of gas at $\text{STP}$, which of course we must treat as an ideal gas to get the molar quantity:

$n = \frac{P V}{R T} = \left(1 \cdot \cancel{\text{bar"xx10^3*cancelL)/(8.31xx10^-2*cancel"bar}} \cdot \cancel{{K}^{- 1}} \cdot m o {l}^{- 1} \cdot \cancel{L} \times 273 \cdot \cancel{K}\right)$

$= 44 \cdot \frac{1}{m o l} ^ - 1 = 44 \cdot \frac{1}{\frac{1}{m o l}} = 44 \cdot m o l$.

We have $\Delta {H}_{\text{combustion}}^{\circ} = - 2220 \cdot k J \cdot m o {l}^{-} 1$, and so we multiply the molar quantity by $\Delta {H}_{\text{combustion}}^{\circ}$

$= 44 \cdot m o l \times - 2220 \cdot k J \cdot m o {l}^{-} 1 \cong {10}^{5} \cdot k J$

Note that I have made assumptions with respect to $\text{standard conditions}$, which might be different to those specified by your syllabus; the UK syllabus is different to IB is different to the USA syllabus. I have ASSUMED, $\text{1 bar pressure}$, and $273 \cdot K$.