# What mass of hydrogen is required to produce 10,000 kJ of energy?

Dec 19, 2014

The answer is $8.3$ $g$ of ${H}_{2}$.

${H}_{2 \left(g\right)} + \frac{1}{2} {O}_{2 \left(g\right)} \to {H}_{2} {O}_{\left(g\right)}$

Here are the standard enthalpy of formation values for the reactants and the products:

${H}_{2 \left(g\right)}$ - $\Delta {H}_{f}^{\circ} = 0$ $\frac{k J}{m o l e}$;
${O}_{2 \left(g\right)}$ - $\Delta {H}_{f}^{\circ} = 0$ $\frac{k J}{m o l e}$;
${H}_{2} {O}_{\left(g\right)}$ - $\Delta {H}_{f}^{\circ} = - 241.8$ $\frac{k J}{m o l e}$.

We know that, for this reaction, the change in enthalpy, $\Delta H$, is $- 241.8$ $k J$; we also know that $\Delta H$ is defined as the sum of the standard enthalpies of formation ($\Delta {H}_{f}^{\circ}$) of the products minus the sum of the standard enthalpies of formation ($\Delta {H}_{f}^{\circ}$) of the reactans - each multiplied by their stoichiometric coefficients.

Therefore, we could write

$\Delta H = 1 m o l e \cdot {\left(\Delta {H}_{f}^{\circ}\right)}_{{H}_{2} O} - \left(0.5 m o l e s \cdot {\left({\Delta}_{f}^{\circ}\right)}_{{O}_{2}} + 1 m o l e \cdot {\left({\Delta}_{f}^{\circ}\right)}_{{H}_{2}}\right) = {\left(\Delta {H}_{f}^{\circ}\right)}_{{H}_{2} O} = - 241.8 k J$

This means that the quantity of ${H}_{2}$ used in the reaction will impact $\Delta H$ through the mole-to-mole ratio it has with ${H}_{2} O$. So, since we know that $\Delta H = - q$, we can determine that we need a new $\Delta H$ of

$\Delta {H}_{n e w} = - q = - 1.00 \cdot {10}^{3} k J$,

which is bigger (in absolute terms) than the original $\Delta H$. This means that we have more moles of ${H}_{2}$ reacting to produce more moles of ${H}_{2} O$. Let's set up the equation with $x$ moles of ${H}_{2} O$, instead of 1:

$\Delta {H}_{n e w} = \left(x m o l e s\right) \cdot {\left(\Delta {H}_{f}^{\circ}\right)}_{{H}_{2} O} \to x = \frac{\Delta {H}_{n e w}}{\Delta {H}_{f}^{\circ}} _ \left({H}_{2} O\right)$, so

$x = \frac{- 1.00 \cdot {10}^{3} k J}{- 241.8 k J} = 4.14$ moles of ${H}_{2} O$ need to be produced in order to release that amount of energy.

Since we know that ${H}_{2}$'s molar mass is $2.0 \frac{g}{m o l}$, and that 1 mole of ${H}_{2}$ produces 1 mole of ${H}_{2} O$, we get

${n}_{{H}_{2}} = 4.14$ moles $\to {m}_{{H}_{2}} = n \cdot m o l a r m a s s = 4.14 m o l e s \cdot 2.0 \frac{g}{m o l} = 8.3 g$

Dec 19, 2014

8.27 g are required.

${H}_{2} + \frac{1}{2} {O}_{2} \rightarrow {H}_{2} O$

$\Delta H = - 241.8 k J$

So:

241.8kJ are produced from I mole = 2g ${H}_{2}$

So:

1kJ produced from $\frac{2}{241.8} g$

So 1000kJ produced from $\frac{2}{241.8} \times 1000 = 8.27 g$