Enthalpy calculations with amount of substance

What ΔrH actually refers to

• A quoted $\Delta_rH$ applies to the exact amounts shown in the balanced equation.

$$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \qquad \Delta_rH = -890\ kJ\ mol^{-1}$$

• This means 890 kJ is released when 1 mol of $CH_4$ burns.
• React a different amount and the energy scales in proportion.

$$\text{energy} = n \times \Delta_rH$$

• n is the number of moles of the substance the $\Delta_rH$ refers to.

Finding the number of moles

The resource booklet gives you the relationships; the skill is choosing the right one.

• From a mass: $n = \frac{m}{M}$

  • n — amount, in mol
  • m — mass, in g
  • M — molar mass, in $g\ mol^{-1}$ (add up the atomic masses from the periodic table)

• From a solution: $n = cV$

  • c — concentration, in $mol\ L^{-1}$
  • V — volume, in L (divide millilitres by 1000)

The method

• Write the balanced equation with its $\Delta_rH$.
• Identify which substance the $\Delta_rH$ refers to (usually the one with a coefficient of 1).
• Calculate n for that substance.
• Multiply: energy $= n \times \Delta_rH$.
• Check the sign, and give the answer to 3 significant figures with units.

Using the mole ratio

• If the amount you are given is for a different substance, convert using the ratio in the equation first.
• Example: in $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$, 2 mol of $O_2$ react per 1 mol of $CH_4$. So 0.40 mol of $O_2$ corresponds to $0.40 \div 2 = 0.20$ mol of $CH_4$, and it is that 0.20 mol you multiply by $\Delta_rH$.

Watching the units

• Mass in grams, not kilograms.
• Volume in litres for $n = cV$ — 250 mL is 0.250 L.
• $\Delta_rH$ in $kJ\ mol^{-1}$, so the energy comes out in kJ.