Question

A scientist measures the standard enthalpy change for the following reaction to be -213.5 kJ: CO(g) 3 H2(g)CH4(g) H2O(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H2O(g) is kJ/mol. Submit AnswerRetry Entire Group

Answer 1

Answer:

Approximately $\rm -249.4\; kJ \cdot mol^{-1}$.

Explanation:

$\rm CO\; (g) + 3\; H_2\; (g) \to CH_4\; (g) + H_2 O\; (g)$.

Note that hydrogen gas $\rm H_2\; (g)$ is the most stable allotrope of hydrogen. Since $\rm H_2$ is naturally a gas under standard conditions, the standard enthalpy of formation of $\rm H_2\; (g)$ would be equal to zero. That is:

• $\Delta H^{\circ}_f(\rm H_2\; (g)) = 0$

Look up the standard enthalpy of formation for the other species:

• $\Delta H^{\circ}_f(\rm CO\; (g)) = -110.5\; kJ \cdot mol^{-1}$,
• $\Delta H^{\circ}_f(\rm CH_4\; (g)) = -74.6\; kJ \cdot mol^{-1}$.

(Source: CRC Handbook of Chemistry and Physics, 84th Edition (2004).)

$\displaystyle \Delta H^{\circ}_\text{reaction} = \sum \Delta H^{\circ}_f(\text{products}) - \sum \Delta H^{\circ}_f(\text{reactants})$.

In other words, the standard enthalpy change of a reaction is equal to:

• the sum of enthalpy change of all products, minus
• the sum of enthalpy change of all reactants.

In this case,

$\begin{aligned} & \sum \Delta H^{\circ}_f(\text{products}) \\ =& \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + \Delta H^{\circ}_f(\mathrm{H_2O\;(g)})\end{aligned}$.

$\begin{aligned} & \sum \Delta H^{\circ}_f(\text{reactants}) \\ =& \Delta H^{\circ}_f(\mathrm{CO\;(g)}) + 3\times \Delta H^{\circ}_f(\mathrm{H_2\;(g)})\end{aligned}$.

Note that the number $3$ in front of $\Delta H^{\circ}_f(\mathrm{H_2\;(g)})$ corresponds to the coefficient of $\rm H_2$ in the chemical equation.

$\begin{aligned}&\Delta H^{\circ}_\text{reaction} \\ =& \sum \Delta H^{\circ}_f(\text{products}) - \sum \Delta H^{\circ}_f(\text{reactants})\\ =& \left(\Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + \Delta H^{\circ}_f(\mathrm{H_2O\;(g)})\right) \\ &- \left(\Delta H^{\circ}_f(\mathrm{CO\;(g)}) + 3\times \Delta H^{\circ}_f(\mathrm{H_2\;(g)})\right) \\ =& \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + (-74.6) - (3 \times 0 -110.5)\end{aligned}$.

In other words,

$\begin{aligned} & \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + (-74.6) - (3 \times 0 -110.5) \\=& \Delta H^{\circ}_\text{reaction} = -213.5\; \rm kJ\cdot mol^{-1} \end{aligned}$.

Therefore,

$\begin{aligned}& \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) \\ =& -213.5 - ((-74.6) - (3 \times 0 -110.5)) \\=& -249.4\; \rm kJ\cdot mol^{-1} \end{aligned}$.