ThermochemistryHard

Question

As a 0.1 mole sample of solid NH4Cl was dissolved in 50 ml of water, the temperature of the solution decreased. A small electrical immersion heater restored the temperature of the system by passing 0.125 A from a 15 V power supply for a period of 14 min. The value of ΔH for the process NH4Cl(s) → NH4Cl(aq) is

Options

A.−15.75 kJ
B.+15.75 kJ
C.−787.5 J
D.+787.5 J

Solution

$q = v.i.t = 15 \times 0.125 \times (14 \times 60)\text{ J = 1575 J}$

$\therefore\Delta H = \frac{1575}{0.1} \times 1 = 15750\text{ J/mol}$

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