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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

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jacob193

Answer:

Approximately [tex]\rm -249.4\; kJ \cdot mol^{-1}[/tex].

Explanation:

[tex]\rm CO\; (g) + 3\; H_2\; (g) \to CH_4\; (g) + H_2 O\; (g)[/tex].

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

  • [tex]\Delta H^{\circ}_f(\rm H_2\; (g)) = 0[/tex]

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

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

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

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

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,

[tex]\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}[/tex].

[tex]\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}[/tex].

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

[tex]\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}[/tex].

In other words,

[tex]\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}[/tex].

Therefore,

[tex]\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}[/tex].

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