JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
Total enthalpy change for freezing of 1 mol of water at \(10^{\circ} \mathrm{C}\) to ice at \(-10^{\circ} \mathrm{C}\) is _______.
(Given : \(\Delta_{\text {fus }} H=x \mathrm{~kJ} / \mathrm{mol}\))
\(\begin{aligned} & \mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_2 \mathrm{O}(\mathrm{l})\right]=\mathrm{y} \mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1} \\ & \mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_2 \mathrm{O}(\mathrm{s})\right]=\mathrm{z} \mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\end{aligned}\)
- A \(-x-10 y-10 z\)
- B \(-10(100 x+y+z)\)
- C \(10(100 x+y+z)\)
- D \(x-10 y-10 z\)
Answer & Solution
Correct Answer
(B) \(-10(100 x+y+z)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \Delta H=1 \times y(0-10)-x \times 1000+1 \times z\left(-10^{\circ}-0^{\circ}\right) \\ & \Delta H=-10(100 x+y+z) \text { Joule. }\end{aligned}\)
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