JEE Main 2025 Apr 7 Shift 1, Chemistry Q4: Hess’s Law and Bond Energy

JEE Main2025Apr 7, Shift 1Chemistry

Total enthalpy change for freezing of 1 mol of water at 10 $^{\circ}$ C to ice at $- 1 0^{\circ}$ C is _____

(Given : $Δ_{fus} H = x$ kJ/mol

$C_{p} [H_{2} O (l)] = y$ J mol $^{- 1}$ K $^{- 1}$

$C_{p} [H_{2} O (s)] = z$ J mol $^{- 1}$ K $^{- 1}$

A
$- x - 10 y - 10 z$
B
$- 10 (100 x + y + z)$
C
$10 (100 x + y + z)$
D
$x - 10 y - 10 z$

Solution

Break the overall freezing process into three reversible steps and sum the enthalpy changes:

Step 1: Cool liquid water from 10 $^{\circ}$ C to 0 $^{\circ}$ C.

$Δ H_{1} = n C_{p} (l) Δ T = (1) (y) (0 - 10) = - 10 y$ J

Step 2: Freeze liquid water at 0 $^{\circ}$ C to ice at 0 $^{\circ}$ C. Since $Δ_{fus} H = x$ kJ/mol refers to melting, freezing releases the same magnitude with a negative sign.

$Δ H_{2} = - x \times 1000$ J

Step 3: Cool ice from 0 $^{\circ}$ C to $- 1 0^{\circ}$ C.

$Δ H_{3} = n C_{p} (s) Δ T = (1) (z) (- 10 - 0) = - 10 z$ J

Summing:

$Δ H = - 10 y - 1000 x - 10 z = - 10 (100 x + y + z)$ J

$∴$ Option (2).

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JEE Main 2025 Apr 7 Shift 1, Chemistry Q4: Hess’s Law and Bond Energy

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