Given:
C(graphite) + O$$_{2}$$(g) → CO$$_{2}$$(g); $$\Delta_{r}$$H° = -393.5 kJ mol$$^{-1}$$
H$$_{2}$$(g) + $$\frac{1}{2}$$O$$_{2}$$(g) → H$$_{2}$$O(l); $$\Delta_{r}$$H° = -285.8 kJ mol$$^{-1}$$
CO$$_{2}$$(g) + 2H$$_{2}$$O(l) → CH$$_{4}$$(g) + 2O$$_{2}$$(g); $$\Delta_{r}$$H° = +890.3 kJ mol$$^{-1}$$
Based on the above thermochemical equations, the value of $$\Delta_{r}$$H° at 298 K for the reaction
C(graphite) + 2H$$_{2}$$(g) → CH$$_{4}$$(g) will be:

We have to evaluate the standard enthalpy change for the reaction

$$C(\text{graphite}) + 2H_2(g) \rightarrow CH_4(g)$$

To do this we shall use Hess’s law, which states: if a chemical equation can be obtained by algebraic addition of two or more other equations, the corresponding enthalpy change is the algebraic sum of the enthalpy changes of the individual steps.

The three thermochemical equations supplied are

$$\text{(1)}\;\;C(\text{graphite}) + O_2(g) \rightarrow CO_2(g),\;\; \Delta_rH_1^\circ = -393.5\ \text{kJ mol}^{-1}$$

$$\text{(2)}\;\;H_2(g) + \dfrac{1}{2}O_2(g) \rightarrow H_2O(l),\;\; \Delta_rH_2^\circ = -285.8\ \text{kJ mol}^{-1}$$

$$\text{(3)}\;\;CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g),\;\; \Delta_rH_3^\circ = +890.3\ \text{kJ mol}^{-1}$$

Our target reaction contains two moles of hydrogen gas, so we first multiply equation (2) by 2. When we multiply an equation by a factor, we must multiply the enthalpy change by the same factor.

$$\text{(2)}\times 2:\;\;2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$$ $$\Delta_rH_2^\circ(\text{new}) = 2 \times (-285.8) = -571.6\ \text{kJ mol}^{-1}$$

Next we add equation (1) and the new version of equation (2). Term by term we get

$$C(\text{graphite}) + O_2(g) \rightarrow CO_2(g)$$

$$2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$$

Adding gives

$$C(\text{graphite}) + 2H_2(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$$

The combined enthalpy change is

$$\Delta H_{\text{sum}}^\circ = -393.5\ \text{kJ mol}^{-1} + (-571.6\ \text{kJ mol}^{-1}) = -965.1\ \text{kJ mol}^{-1}$$

Now we introduce equation (3) exactly as it is:

$$CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g)$$

with

$$\Delta_rH_3^\circ = +890.3\ \text{kJ mol}^{-1}$$

We add this equation to the sum obtained previously. Writing both together for clarity:

$$C(\text{graphite}) + 2H_2(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$$

$$CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g)$$

On adding, identical species appearing on opposite sides cancel:

• $$CO_2(g)$$ appears once on the right and once on the left  →  cancels.
• $$2H_2O(l)$$ appears once on the right and once on the left  →  cancels.
• $$2O_2(g)$$ appears once on the left and once on the right  →  cancels.

The remaining, uncancelled equation is precisely the target:

$$C(\text{graphite}) + 2H_2(g) \rightarrow CH_4(g)$$

According to Hess’s law the enthalpy change for the desired reaction is therefore

$$\Delta_rH^\circ = \Delta H_{\text{sum}}^\circ + \Delta_rH_3^\circ$$

$$\Delta_rH^\circ = (-965.1\ \text{kJ mol}^{-1}) + (+890.3\ \text{kJ mol}^{-1})$$

$$\Delta_rH^\circ = -74.8\ \text{kJ mol}^{-1}$$

Hence, the correct answer is Option B.