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JEE Main Chemistry Thermodynamics (C) 2026 JEE Main 2026 (28 January Shift 2)

JEE Main Chemistry Question (2026) — Solution

Question

The plot of _ 10 ~K vs 1 ~T gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).

Options

  1. A. S ^ o 2 303 R ,- H ^ o 2 303 R
  2. B. - S ^ R 2 303 , H ^ R 2 303
  3. C. 2 303 R H ^ o , 2 303 R ~S ^ o
  4. D. - H ^ o 2 303 R , ~S ^ o 2 303 R

Answer

A. S ^ o 2 303 R ,- H ^ o 2 303 R

Step-by-step solution

The relationship between the standard Gibbs free energy change and the equilibrium constant is given by G^o = -RT K. We also know that G^o = H^o - T S^o. Equating the two expressions: -RT K = H^o - T S^o. Dividing by -RT, we get: K = - H^o RT + S^o R . To convert natural log to base 10, use K = 2.303 _ 10 K: 2.303 _ 10 K = - H^o RT + S^o R . Dividing by 2.303: _ 10 K = - H^o 2.303RT + S^o 2.303R . Rearranging in the form of a straight line equation y = mx + c where y = _ 10 K and x = 1 T : _ 10 K = ( - H^o 2.303R ) 1 T + S^o 2.303R . Comparing the terms: Intercept (c) = S^o 2.303R Slope (m) = - H^o 2.303R .

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