# Use Hess's Law and the following information to calculate the change in enthalpy for the reaction 2C + H2 -> C2H2?

## C2H2 + 5/2 O2 -> 2CO2 + H2O Entropy = -1299.6 kJ C + O2 -> CO2 Entropy = -393.5 kJ H2 + 1/2 O2 -> H2O Entropy = -285.8 kJ

Oct 15, 2016

$2 C \left(s\right) + {H}_{2} \left(g\right) \rightarrow H C \equiv C H \left(g\right)$
$\Delta {H}^{\circ} \text{_"reaction} = 227 \cdot k J \cdot m o {l}^{-} 1$

#### Explanation:

You have listed entropy changes for the reaction. In fact, these are values for $\Delta {H}^{\circ} \text{_"reaction}$

$H C \equiv C H + \frac{5}{2} {O}_{2} \rightarrow 2 C {O}_{2} + {H}_{2} O$ ;DeltaH^@""_"reaction"=-1299.6*kJ*mol^-1 $\left(i\right)$

$C + {O}_{2} \rightarrow C {O}_{2}$ ;DeltaH^@""_"reaction"=-393.5*kJ*mol^-1 $\left(i i\right)$

${H}_{2} + \frac{1}{2} {O}_{2} \rightarrow {H}_{2} O$ ;DeltaH^@""_"reaction"=-285.6*kJ*mol^-1 $\left(i i i\right)$

We want $\Delta {H}^{\circ}$ for:

$2 C + {H}_{2} \rightarrow H C \equiv C H$ ;DeltaH^@=??

If I do the sum of the given reactions in this manner, $2 \times \left(i i\right) - \left(i\right) + \left(i i i\right)$, I get the following:

$2 C + \cancel{2 {O}_{2}} + \cancel{2 C {O}_{2}} + \cancel{{H}_{2} O} + {H}_{2} + \cancel{\frac{1}{2} {O}_{2}} \rightarrow \cancel{2 C {O}_{2}} + H C \equiv C H + \cancel{{H}_{2} O} + \cancel{\frac{5}{2} {O}_{2}}$

Which after cancelling gives:

$2 C \left(s\right) + {H}_{2} \left(g\right) \rightarrow H C \equiv C H \left(g\right)$ $\Delta {H}^{\circ} \text{_"reaction} = 227 \cdot k J \cdot m o {l}^{-} 1$

Note here that $\Delta {H}^{\circ} \text{_"reaction"=DeltaH^@""_f" acetylene}$.
So here we have used the thermochemical equations as linear equations to give us the equation that we sought (and thus necessarily the thermodynamic parameters).