JEE Main 2026 — Jan 28, Shift 2 — Chemistry solution

JEE Main2026Jan 28, Shift 2Chemistry

The plot of $lo g_{10} K$ vs $\frac{1}{T}$ gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).

Solution

From $Δ G^{\circ} = - R T ln K$ and $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ :

$lo g_{10} K = - \frac{Δ H ^{\circ}}{2.303 R} \cdot \frac{1}{T} + \frac{Δ S ^{\circ}}{2.303 R}$ .

Comparing with $y = m x + c$ where $x = 1/ T$ :

Slope $= - \frac{Δ H ^{\circ}}{2.303 R}$ , $y$ -intercept $= \frac{Δ S ^{\circ}}{2.303 R}$ .

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