# Question 6ecfc

Mar 20, 2015

The standard enthalpy of reaction for ethane, or ${C}_{2} {H}_{6}$, is $\text{-84.6 kJ/mol}$.

So, start by writing your balanced chemical equations again

(1)$\text{ } {C}_{\left(s\right)} + {O}_{2 \left(g\right)} \to C {O}_{2 \left(g\right)}$, DeltaH_("rxn 1")^(0) = "-393.5 kJ/mol"

(2)$\text{ "H_(2(g)) + "1/2} {O}_{2 \left(g\right)} \to {H}_{2} {O}_{\left(l\right)}$, DeltaH_("rxn 2")^(0) = "-285.8 kJ/mol"

(3)$\text{ } 2 {C}_{2} {H}_{6 \left(g\right)} + 7 {O}_{2 \left(g\right)} \to 4 C {O}_{2 \left(g\right)} + 6 {H}_{2} {O}_{\left(l\right)}$, DeltaH_("rxn 3")^(0) = "-3119.6 kJ/mol"

(4)$\text{ } 2 {C}_{\left(s\right)} + 3 {H}_{2 \left(g\right)} \to {C}_{2} {H}_{6 \left(g\right)}$, DeltaH_("rxn 4")^(0) = "?"

To solve this problem you'll have to use Hess' law, which states that the enthalpy of a chemical reaction is constant regardless of the number of steps in which said reaction takes place.

The standard enthalpy of reaction is defined as the difference between the sum of the standard enthalpies of formation of the products and the sum of the standard enthalpies of formation of the reactants.

$\Delta {H}_{\text{rxn")^(0) = Sigma(DeltaH_f^(0) "products") - Sigma(DeltaH_f^(0)"rectants}}$

So, all you need to actually do is figure out how you can use the enthalpies of reaction given to write the enthalpy of reaction for the formation of ethane.

Start with equation (1). Since the standard enthalpies of formation for elements in their most stable natural state is zero, you can write

$\Delta {H}_{\text{rxn 1}} = \Delta {H}_{f}^{0} \left(C {O}_{2}\right) - \left(0 + 0\right)$

DeltaH_f^(0)(CO_2) = DeltaH_("rxn 1") = "-393.5 kJ/mol"

On to equation (2). The same approach is valid in this case as well, which means that

DeltaH_f^(0)(H_2O) = DeltaH_("rxn 2") = "-285.8 kJ/mol"

So, you know the enthalpy of formation for 1 mole of carbon dioxide gas and 1 mole of water.

If you look closely at equation (3), you'll notice that you can use these standard enthalpies of formation to determine the standard enthalpy of formation for 2 moles of ethane.

You must however multiply these standard enthalpies of formation by the number of moles of each substance that takes part in the reaction - 4 moles for $C {O}_{2}$ and 6 moles for ${H}_{2} O$.

$\Delta {H}_{\text{rxn 3}} = \left(4 \cdot \Delta {H}_{f}^{0} \left(C {O}_{2}\right) + 6 \cdot \Delta {H}_{f}^{0} \left({H}_{2} O\right)\right) - \left(2 \cdot \Delta {H}_{f}^{0} \left({C}_{2} {H}_{6}\right)\right)$

Solving for $\Delta {H}_{f}^{0} \left({C}_{2} {H}_{6}\right)$ will give you

$2 \cdot \Delta {H}_{f}^{0} \left({C}_{2} {H}_{6}\right) = \left(4 \cdot \Delta {H}_{f}^{0} \left(C {O}_{2}\right) + 6 \cdot \Delta {H}_{f}^{0} \left({H}_{2} O\right)\right) - \Delta {H}_{\text{rxn 3}}$

2 * DeltaH_f^(0)(C_2H_6) = (4 * ("-393.5 kJ/mol") + 6 * ("-285.8 kJ/mol") - ("-3119.6 kJ/mol")

$2 \cdot \Delta {H}_{f}^{0} \left({C}_{2} {H}_{6}\right) = \text{-169.2 kJ/mol}$

This means that

$\Delta {H}_{f}^{0} \left({C}_{2} {H}_{6}\right) = \text{-169.2 kJ/mol"/2 = "-84.6 kJ/mol}$

As a result, $\Delta {H}_{\text{rxn 4}}$ will be

DeltaH_("rxn 4") = DeltaH_f^(0)(C_2H_6) = "-84.6 kJ/mol"#