JEE MAIN - Chemistry (2025 - 7th April Morning Shift - No. 3)
Total enthalpy change for freezing of 1 mol of water at $10^{\circ} \mathrm{C}$ to ice at $-10^{\circ} \mathrm{C}$ is ________
(Given: $\Delta_{\text {fus }} \mathrm{H}=x \mathrm{~kJ} / \mathrm{mol}$
$$\begin{aligned} & \mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_2 \mathrm{O}(\mathrm{l})\right]=y \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\ & \mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_2 \mathrm{O}(\mathrm{~s})\right]=z \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \end{aligned}$$
$-x-10 y-10 z$
$x-10 y-10 z$
$-10(100 x+y+z)$
$10(100 \mathrm{x}+y+z)$
Explanation
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$$\begin{aligned} & \Delta H=1 \times y(0-10)-x \times 1000+1 \times z\left(-10^{\circ}-0^{\circ}\right) \\ & \Delta H=-10(100 x+y+z) \text { Joule. } \end{aligned}$$
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