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

Approximately
`\rm -249.4\; kJ \cdot mol^(-1)`.

Step-by-step 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* \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* \Delta H^(\circ)_f(\mathrm{H_2\;(g)})\right) \\ =& \Delta H^(\circ)_f(\mathrm{CH_4\;(g)}) + (-74.6) - (3 * 0 -110.5)\end{aligned}`.

In other words,

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