Questions 39-41 refer to the following graph, which shows the heating curve for methane, CH4. [VISUA — Thermodynamics Chemistry Question

STEPS:

1. Convert the mass of methane to moles: Use the mass of the methane sample ($64\text{ g}$) and its molar mass ($16\text{ g/mol}$) to calculate the total moles of methane being heated:
$$\text{Moles of }\text{CH}_4 = \frac{64\text{ g}}{16\text{ g/mol}} = \mathbf{4.0\text{ mol}} \quad$$
2. Identify the phase change on the heating curve: Melting (fusion) is a constant-temperature phase transition from solid to liquid. On the provided heating curve, this process occurs at the first horizontal plateau (labeled Q) at a temperature of $90\text{ K}$.
3. Retrieve the molar enthalpy of fusion ($\Delta H_{\text{fus}}$): The plateau Q is annotated with a molar enthalpy of fusion value:
$$\Delta H_{\text{fus}} = \mathbf{0.94\text{ kJ/mol}} \quad$$
This constant tells us that $0.94\text{ kJ}$ of heat energy must be absorbed to melt exactly one mole of solid methane at $90\text{ K}$.
4. Calculate the total heat energy required: Multiply the moles of methane by the molar enthalpy of fusion:
$$q = n \times \Delta H_{\text{fus}}$$
$$q = 4.0\text{ mol} \times 0.94\text{ kJ/mol} = \mathbf{3.76\text{ kJ}}$$
Rounding to two significant figures yields $3.8\text{ kJ}$, which corresponds to Option B.

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WHY_OTHERS_WRONG:

* Option A is incorrect: The value $0.24\text{ kJ}$ is calculated if a student divides the molar enthalpy of fusion by the number of moles ($0.94\text{ kJ/mol} / 4.0\text{ mol} \approx 0.24\text{ kJ}$) rather than multiplying them.
* Option C is incorrect: The value $33\text{ kJ}$ is obtained if a student mistakenly uses the molar enthalpy of vaporization ($\Delta H_{\text{vap}} = 8.2\text{ kJ/mol}$) instead of the enthalpy of fusion:
$$q = 4.0\text{ mol} \times 8.2\text{ kJ/mol} = 32.8\text{ kJ} \approx 33\text{ kJ}$$
Vaporization represents boiling (the liquid-to-gas transition at the higher plateau S at $110\text{ K}$), not melting at $90\text{ K}$.
* Option D is incorrect: The value $60.\text{ kJ}$ is produced if a student multiplies the mass directly by the enthalpy of fusion ($64\text{ g} \times 0.94\text{ kJ/mol} \approx 60.\text{ kJ}$) without first converting grams to moles. This calculation is incorrect because the unit of enthalpy of fusion is per *mole*, not per *gram*.