JEE Main 2017 · 8 Apr · Shift 1 · Chemistry Q40

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ChemistrySingle correct+4 / -1Question 40 of 90

The enthalpy change on freezing of 1 mol of water at 5°C to ice at −5°C is :

(Given $\Delta_{fus}H=6$ kJ mol $^{-1}$ at 0°C, $C_p(\ce{H2O}, l)=75.3$ J mol $^{-1}$ K $^{-1}$ , $C_p(\ce{H2O}, s)=36.8$ J mol $^{-1}$ K $^{-1}$ )

5.44 kJ mol

^{-1}

5.81 kJ mol

^{-1}

6.56 kJ mol

^{-1}

6.00 kJ mol

^{-1}

Answer and worked solution

Official answer: C

Three steps: cooling water 5 → 0 °C gives $-75.3\times5=-376.5$ J; freezing gives $-6000$ J; cooling ice 0 → −5 °C gives $-36.8\times5=-184$ J. In total $\Delta H=-6.56$ kJ mol⁻¹: the options give the magnitude, 6.56 kJ mol⁻¹.

Topics:Hess's law and enthalpies of reaction Calorimetry

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JEE Main 2017 · 8 Apr · Shift 1, Chemistry question 40