Plot of log10 K vs 1 by T.... | JEE Main 2026 28 Jan Shift 2

JEE MAIN_2026_

280126_S2

Question

The plot of $\log _{10} \mathrm{~K} v s \frac{1}{\mathrm{~T}}$ gives a straight line. The intercept and slope respectively are (where $K$ is equilibrium constant).

Select the correct option:

$-\frac{\Delta \mathrm{S}^{\circ} \mathrm{R}}{2 \cdot 303}, \frac{\Delta \mathrm{H}^{\circ} \mathrm{R}}{2 \cdot 303}$

$\frac{\Delta \mathrm{S}^{\circ}}{2 \cdot 303 \mathrm{R}},-\frac{\Delta \mathrm{H}^{\circ}}{2 \cdot 303 \mathrm{R}}$

$-\frac{\Delta \mathrm{H}^{\circ}}{2 \cdot 303 \mathrm{R}}, \frac{\Delta \mathrm{S}^{\circ}}{2 \cdot 303 \mathrm{R}}$

$\frac{2 \cdot 303 \mathrm{R}}{\Delta \mathrm{H}^{\circ}}=\frac{2 \cdot 303 \mathrm{R}}{\Delta \mathrm{S}^{\circ}}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Chemistry

Easy

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