Answer :
Final answer:
The amount of heat released when 8.2 mol of butane freezes can be calculated using the equation q = m * ΔH. The correct answer is none of the options given in the question.
Explanation:
When a substance freezes, heat is released. The amount of heat released can be calculated using the equation q = m * ΔH, where q is the heat released, m is the mass of the substance, and ΔH is the enthalpy change per mole of the substance during freezing. Given that butane has a molar mass of 58.12 g/mol and its enthalpy change of fusion is -125.7 kJ/mol, we can calculate the amount of heat released when 8.2 mol of butane freezes:
q = 8.2 mol * -125.7 kJ/mol
q = -125.7 kJ/mol * 8.2 mol
q = -1030.14 kJ
Converting kJ to kilocalories:
1 kcal = 4.184 kJ
-1030.14 kJ * (1 kcal / 4.184 kJ) = -246.42 kcal
The amount of heat released when 8.2 mol of butane freezes is approximately 246.42 kilocalories. Therefore, none of the given options a), b), c), or d) are correct.
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