# Enthalpy Calculations, Heat of Formation/Combustion

I just don't understand how to do these and would really appreciate your help.

1) Ammonium dichromate decomposes in a reaction when heated. Calculate the heat transferred for the decomposition of 53.0 g of ammonium dichromate according to $\ce{(NH4)2Cr2O7(s) -> Cr2O3(s) + N2(g) + 4H2O(g)}$ ΔH° = -315 kJ Do I convert grams to moles and then multiply by ΔH° ?

2) The standard heat of formation $\Delta H_f$ for $\ce{C2H4}$ is +52.3 kJ/mol. If $\ce{C2H4}$ (ethene) reacts with $\ce{H2}$(g) to produce $\ce{C2H6}$ (g) (ethane) ΔH = -137 kJ. What is the heat formation of ethane?

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Thanks for at least hinting at your approach for number 1, can you give a similar statement for number 2? –  jonsca Feb 11 '13 at 0:00

Exercise 1

The decomposition of ammonium dichromate has a standard enthalpy change of $-315 \:\mathrm{kJ/mol}$. So you have a figure that relates the heat to moles. You need to find out how many moles correspond to 53.0 g of the ammonium dichromate. We use molar mass for this. You can calculate molar mass by simply summing the weights of all atoms in the molecule, but I use this tool to do it. $$\ce{M}_{\ce{(NH4)2Cr2O7}}=252.06 \:\mathrm{\frac{g}{mol}}$$

$$\:\mathrm{g/\frac{g}{mol}=mol}$$ $$\Downarrow$$ $$\frac{53.0\:\mathrm{g}}{252.06 \:\mathrm{\frac{g}{mol}}}=0.21\:\mathrm{mol}$$

So far so good. Our figure of enthalpy change is per mole, thus you simply multiply the two:

$$\:\mathrm{\frac{kJ}{mol}*mol=kJ}$$ $$\Downarrow$$ $$\Delta H^\Theta-315 \:\mathrm{\frac{kJ}{mol}}*0.21\:\mathrm{mol}=-66.15\:\mathrm{kJ}$$

Exercise 2

To solve this exercise you need to apply Hess' law:

$$\Delta H^\Theta_{reaction}=\sum \Delta H^\Theta _{f(products)}-\sum \Delta H^\Theta _{f(reactants)}$$

So we actually need a figure for hydrogen. You look that up on the internet for example, and see that $\Delta H^\Theta _f \ce{(H2)}=0 \:\mathrm{\frac{kJ}{mol}}$.

We now use Hess' law. We have the enthalpy of the reaction, and we have the formation enthalpies of the reactants. So we rearrange the equation to find the formation enthalpy for the single product. Notice we leave out hydrogen only because it's 0:

$$\Delta H^\Theta_{reaction}=\Delta H^\Theta _{f(ethane)}-\Delta H^\Theta _{f(ethene)}$$ $$\Downarrow$$ $$\Delta H^\Theta _{f(ethane)} = \Delta H^\Theta_{reaction}+\Delta H^\Theta _{f(ethene)}$$ $$\Delta H^\Theta _{f(ethane)} = -137\:\mathrm{\frac{kJ}{mol}} + 53.3\:\mathrm{\frac{kJ}{mol}}=-83.7\:\mathrm{\frac{kJ}{mol}}$$

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