MolSynapsequestion bank

MolSynapse / Question bank / Chemical Equilibrium / Thermodynamics of equilibrium: van't Hoff equation

JEE Main levelSingle correct optionDifficulty 2 of 3ce136

Take $R = 8.314 J mo l^{- 1} K^{- 1}$ . The equilibrium constant of a reaction is $6.0$ at $400 K$ and $2.4$ at $500 K$ . Assuming $Δ H^{\circ}$ is constant, its value is closest to

A $+ 15.2 kJ mo l^{- 1}$
B $- 6.6 kJ mo l^{- 1}$
C $- 15.2 kJ mo l^{- 1}$
D $+ 6.6 kJ mo l^{- 1}$

Answer: C

ln (K_{2} / K_{1}) = ln (2.4/6.0) = - 0.9163

and

(1/ T_{1} - 1/ T_{2}) = 5.00 \times 1 0^{- 4} K^{- 1}

.

Δ H^{\circ} = R ln (K_{2} / K_{1}) / (1/ T_{1} - 1/ T_{2}) = - 15.2 kJ mo l^{- 1}

.

K

falls with temperature, so the reaction is exothermic; using

lo g_{10}

gives -6.6.

Practice Chemical Equilibrium More on Thermodynamics of equilibrium: van't Hoff equation

Previous questionNext question

More questions on Thermodynamics of equilibrium: van't Hoff equation