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JEE Advanced 2006
Exam
JEE Advanced
Year
2006
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Advanced 2006 · Chemistry · Thermodynamics (C)

Which is correct statement if N_2 is added at equilibrium condition?

Which is correct statement if N_2 is added at equilibrium condition?
Which is correct statement if N2\mathrm{N}_2 is added at equilibrium condition?

Options

  1. AThe equilibrium will shift to forward direction because according to IInd law of thermodynamics the entropy must increases in the direction of spontaneous reaction.
  2. BThe condition for equilibrium is GN2+3GH2=2GNH3G_{\mathrm{N}_2}+3 G_{\mathrm{H}_2}=2 G_{\mathrm{NH}_3} where GG is Gibbs free energy per mole of the gaseous species measured at the partial pressure. The condition of equilibrium is unaffected by the use of catalyst, which increases the rate of both the forward and backward reactions to the same extent.
  3. CThe catalyst will increase the rate of forward reaction by α\alpha and that of backward reaction by β\beta.
  4. DCatalyst will not alter the rate of either of the reaction.

Correct answer

B. The condition for equilibrium is GN2+3GH2=2GNH3G_{\mathrm{N}_2}+3 G_{\mathrm{H}_2}=2 G_{\mathrm{NH}_3} where GG is Gibbs free energy per mole of the gaseous species measured at the partial pressure. The condition of equilibrium is unaffected by the use of catalyst, which increases the rate of both the forward and backward reactions to the same extent.

Step-by-step solution

Which is correct statement if N_2 is added at equilibrium condition?
KporKcK_p or K_c remains constant for above reaction at constant temperature. Although in presence of catalyst rate of forward and backward reaction is increased in same order (extent) due to decreasing the activation energy of both reaction. So the equilibrium is established in short time i.e., catalyst is able to increase the concentration of reactants and products in same order, so the value of KCorKpK_C or K_p remains constant. Hence equilibrium state is not effected in presence of catalyst. Hence, at equilibrium ΔG=0\Delta G=0\begin{aligned}& \because \quad \Delta G=2 G \text{\text{ of } } \mathrm{NH}_{3}-\left(\mathrm{G} \text{ of } \mathrm{H}_{2}+3 \times G \text{ of } \mathrm{NH} }_{3}\right) \\ & \therefore G_{\mathrm{N}_2}+3 G_{\mathrm{H}_2}=2 G_{\mathrm{NH}_3} \end{aligned}$$
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