Explanation
For a spontaneous process the randomness of the molecules in the system increases so the entropy change of the system and surroundings must be positive.
Hence option B is correct.
$$CH_{4}+2O_{2}\rightarrow CO_{2}+2H_{2}O\left ( g \right)+210.8kcal$$
Then the possible heat of formation of methane will be
$$CH_{4}+2O_{2}\rightarrow CO_{2}+2H_{2}O\left ( g \right); \ \Delta H_R=+210.8kcal$$ (iii)
$$\Delta H_R$$ for reaction (iii) can be written as:
$$\Delta H = \Delta H_f (CO_2) +2\times \Delta H_f (H_2O) - \Delta H_F (CH_4) -2\times \Delta H_f (O_2)$$
$$\implies 210.8 = 94.2 + 2 \times 68.3 -\Delta H_f (CH_4) - 2\times 0$$
$$\Delta H_f (CH_4) = 20 \ kcal$$
Hence, Option "B" is the correct answer.
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