Answer:
Approximately [tex]-74.8\; \rm kJ \cdot mol^{-1}[/tex].
Explanation:
Number the three reactions with known enthalpy changes:
[tex]\begin{aligned}& {\rm C\; (s) + O_2\; (g) \to CO_2\; (g)} & \quad \Delta H = -393.5\; \rm kJ\cdot mol^{-1}& \quad (1) \\ & {\rm H_2\; (2) + \frac{1}{2} O_2\; (g) \to H_2O\; (l)} & \quad \Delta H = -285.8\; \rm kJ\cdot mol^{-1} & \quad (2) \\ & {\rm CH_4\; (g) + 2\; O_2\; (g) \to CO_2\; (g) + 2\; H_2O\; (l)} & \quad \Delta H = -890.3\; \rm kJ\cdot mol^{-1} & \quad (3)\end{aligned}[/tex].
The goal is to find a way to combine these three reactions to obtain: [tex]\rm C\; (s) + 2\; H_2\; (g) \to CH_4\; (g)[/tex].
Assume that the three known reactions are combined in this way:
[tex]a \times (1) + b \times (2) + c \times (3)[/tex].
That corresponds to:
[tex]\begin{aligned} & a\; \mathrm{C\; (s)} + \left(a + \frac{1}{2}\, b + c\right) \; \mathrm{O_2\; (g)} + b\; \mathrm{H_2\; (g)} + c\; \mathrm{CH_4\; (g)} \\ & \to (a + c)\; \mathrm{CO_2\; (g)} + (b+ 2\,c )\; \mathrm{H_2O\; (l)}\end{aligned}[/tex].
Compare the coefficients of this reaction with that of the desired reaction:
[tex]\rm C\; (s) + 2\; H_2\; (g) \to CH_4\; (g)[/tex].
Note that some species (e.g., [tex]\rm CH_4\; (g)[/tex]) appeared on the wrong side of the equation. Their desired coefficient should be the opposite of their true coefficient. For example, the coefficient of [tex]\rm CH_4\; (g)[/tex] is supposed to be [tex]1[/tex]. However, because it appeared on the wrong side of the equation, its desired coefficient would be [tex]-1[/tex].
The coefficients of species that are not in the desired equation should be zero.
[tex]\begin{array}{|c|c|c|}\cline{1-3}\text{Species} & \text{Coefficient}& \text{Desired Coefficient} \\ \cline{1-3} \mathrm{C\; (s)}} & a & 1 \\ \cline{1-3} \mathrm{O_2\; (g)} & a + (1/2)\, b + c & 0 \\ \cline{1-3} \mathrm{H_2\; (g)} & b & 2 \\ \cline{1-3}\mathrm{CH_4\; (g)} & c & -1\\ \cline{1-3} \mathrm{CO_2\; (g)} & a + c & 0 \\ \cline{1-3} \mathrm{H_2O\; (l)} & b + 2\, c & 0\\ \cline{1-3}\end{array}[/tex].
Solve for [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex]:
In other words, the desired equation is equal to [tex]1 \times (1) + 2 \times (2) - 1 \times (3)[/tex].
By Hess's Law, the enthalpy of the desired equation will be:
[tex]\begin{aligned}& 1\times \Delta H (1) + 2 \times \Delta H (2) - 1\times \Delta H(3) \\ &\approx -393.5 + 2 * (-285.8) - (-890.3) \\ &= -74.8\; \rm kJ \cdot mol^{-1} \end{aligned}[/tex].
