# What is the ∆H value for this thermo chemical equation? Full question in the description box below.

Jan 10, 2018

$\textsf{- 741 \textcolor{w h i t e}{x} k J \text{/""mol}}$

#### Explanation:

$\textsf{{M}_{r} \left[{C}_{3} {H}_{8}\right] = 44.1}$

We are told that:

1 tonne gives $\textsf{1.68 \times {10}^{4} \textcolor{w h i t e}{x} M J}$

$\therefore$$\textsf{\cancel{{10}^{6}} \textcolor{w h i t e}{x} g \rightarrow 1.68 \times {10}^{4} \times \cancel{{10}^{6}} J}$

This means that:

$\textsf{1 \textcolor{w h i t e}{x} g \rightarrow 1.68 \times {10}^{4} \textcolor{w h i t e}{x} J}$

We need to find the energy released by 1 mole which is 44.1 g:

$\therefore$$\textsf{44.1 \textcolor{w h i t e}{x} g \rightarrow 1.68 \times {10}^{4} \times 44.1 \textcolor{w h i t e}{x} J}$

$\textsf{= 74.088 \times {10}^{4} \textcolor{w h i t e}{x} J}$

$\textsf{= 741 \textcolor{w h i t e}{x} \text{kJ}}$

Since heat is released we can say that:

$\textsf{\Delta H = - 741 \textcolor{w h i t e}{x} \text{kJ""/""mol}}$

Jan 10, 2018

$\Delta H = - 739$ $k J m o {l}^{-} 1$

#### Explanation:

First we need to calculate the number of moles of $C {O}_{2}$ in $1$ tonne, which is ${10}^{6}$ $g$. The molar mass of $C {O}_{2}$ is $44$ $g m o {l}^{-} 1$.

$n = \frac{m}{M} = {10}^{6} / 44 = 22 , 727$ $m o l$

Now the total energy produced by $1$ tonne is $1.68 \times {10}^{4}$ $M J$ = $1.68 \times {10}^{7}$ $k J$. This is energy emitted to the environment, and therefore lost from the chemical system. That means it will take a negative value when we calculate the $\Delta H$.

We need to divide this by the number of moles to find the $\Delta H$ value in $k J m o {l}^{-} 1$.

$\Delta H = - \frac{1.68 \times {10}^{7}}{22 , 727} = - 739$ $k J m o {l}^{-} 1$