The reaction between hydrogen and oxygen to yield water vapor has $\Delta H^\circ = \pu{- 484 kJ}$. How much $pV$ work is done, and what is the value of $\Delta E$ in kilojoules for the reaction of $\pu{0.50 mol}$ of $\ce{H2}$ with $\pu{0.25 mol}$ of $\ce{O2}$ at atmospheric pressure if the volume change is $-\pu{5.6L}?$

$$\ce{2H2(g) + O2(g) -> 2H2O(g)} \qquad \Delta H^\circ = \pu{- 484 kJ}$$

I use the formula $\Delta E=\Delta H - p\Delta V$ to determine $\Delta E.$ However, when determining the enthalpy, the solutions manual does this:

$$\Delta H = \frac{\pu{-121 kJ}}{\pu{0.50 mol}~\ce{H2}}$$

Where does the $-121$ come from? From my understanding, since there are two moles $\ce{H2}$, $\Delta H$ should be $-\pu{242 kJ}$. Or do we take into account all four hydrogen atoms? That would give us $\pu{-484 kJ}/4 = -\pu{121 kJ}.$

  • $\begingroup$ Is ∆E or ∆H change in enthalpy? $\endgroup$
    – Dissenter
    Aug 11, 2014 at 16:58
  • $\begingroup$ $\Delta H$ is change in enthalpy $\endgroup$
    – Amuna
    Aug 11, 2014 at 17:11

1 Answer 1


The confusion is partly caused by careless use of quantities and units. The value $H = -484\ \mathrm{kJ}$ denotes an enthalpy. However, what is actually meant is the enthalpy per one mole; i.e. the molar enthalpy $H_{\mathrm m} = -484\ \mathrm{kJ/mol}$.

In the definition of the molar reaction enthalpy, the ‘per mole’ does not refer to any particular substance in the equation. Instead it refers to the entire reaction as a whole. Therefore, the reaction must be specified for which this quantity applies. In this case, the enthalpy of $484\ \mathrm{kJ}$ is released when $2\ \mathrm{mol}$ of hydrogen gas react with $1\ \mathrm{mol}$ of oxygen gas to form $2\ \mathrm{mol}$ of gaseous water: $$\ce{2H2(g) + O2(g) -> 2H2O(g)}\qquad\Delta H^\circ = -484\ \mathrm{kJ}$$ (By way of comparison, the corresponding value for liquid water is about $-572\ \mathrm{kJ}$.)

Thus, the molar enthalpy relating to the amount of hydrogen is $$\frac{\Delta H^\circ}{n(\ce{H2})}=\frac{-484\ \mathrm{kJ}}{2\ \mathrm{mol}}=\frac{-242\ \mathrm{kJ}}{1\ \mathrm{mol}}=-242\ \mathrm{kJ/mol}$$ i.e. the released enthalpy per $1\ \mathrm{mol}$ of hydrogen gas is $242\ \mathrm{kJ}$.

However, the given question is asking about $n(\ce{H2}) = 0.50\ \mathrm{mol}$ of hydrogen (and accordingly $n(\ce{O2}) = 0.25\ \mathrm{mol}$ of oxygen). The corresponding enthalpy is $$\begin{align} 0.5\ \mathrm{mol}\times\frac{\Delta H^\circ}{n(\ce{H2})}&= 0.5\ \mathrm{mol}\times\frac{-484\ \mathrm{kJ}}{2\ \mathrm{mol}}\\[6pt] &= 0.5\ \mathrm{mol}\times\frac{-242\ \mathrm{kJ}}{1\ \mathrm{mol}}\\[6pt] &= 0.5\ \mathrm{mol}\times -242\ \mathrm{kJ/mol}\\[6pt] &= -121\ \mathrm{kJ} \end{align}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.