JEE Challenger
More from Equilibrium
ChemHardAdv 2020 • Paper 1

Ratio of Standard Gibbs Energy at Different Temperatures from Pressure-Time Plot

For the reversible reaction A⇌B\text{A} \rightleftharpoons \text{B} at 1000 K1000\text{ K}, the temperature of the reaction system is increased to 2000 K2000\text{ K} at time t′t' and allowed to attain a new state of equilibrium. Throughout the experiment, the partial pressure of component A\text{A} is maintained constant at 1 bar1\text{ bar}. Below is the plot displaying the variation of the partial pressure of B\text{B} with respect to time. Calculate the ratio of the standard Gibbs energy change of the reaction at 1000 K1000\text{ K} to that at 2000 K2000\text{ K}.

Question Diagram 1
Official Numerical Answer0.25

Topics & Concepts

EquilibriumChemical EquilibriumGibbs Free Energy

Step-by-Step Solution

For the given reversible reaction: A(g)⇌B(g)\text{A}(g) \rightleftharpoons \text{B}(g)

The equilibrium constant in terms of partial pressures, KpK_p, is expressed as: Kp=PBPAK_p = \frac{P_\text{B}}{P_\text{A}}

Throughout the experiment, the partial pressure of component A\text{A} is maintained constant at: PA=1 barP_\text{A} = 1\text{ bar}

Step 1: Analysis at T1=1000 KT_1 = 1000\text{ K} Before time t′t', the reaction system is at equilibrium at T1=1000 KT_1 = 1000\text{ K}. From the provided plot, the equilibrium partial pressure of B\text{B} is: PB=10 barP_\text{B} = 10\text{ bar}

Thus, the equilibrium constant at 1000 K1000\text{ K} is: Kp,1000=10 bar1 bar=10K_{p, 1000} = \frac{10\text{ bar}}{1\text{ bar}} = 10

The standard Gibbs energy change at 1000 K1000\text{ K} is given by: ΔG∘(1000 K)=−RT1ln⁡(Kp,1000)=−R(1000)ln⁡(10)\Delta G^\circ (1000\text{ K}) = -R T_1 \ln(K_{p, 1000}) = -R (1000) \ln(10)

Step 2: Analysis at T2=2000 KT_2 = 2000\text{ K} After time t′t', the temperature is raised to T2=2000 KT_2 = 2000\text{ K} and the system attains a new state of equilibrium. From the plot, the new equilibrium partial pressure of B\text{B} is: PB=100 barP_\text{B} = 100\text{ bar}

Thus, the equilibrium constant at 2000 K2000\text{ K} is: Kp,2000=100 bar1 bar=100K_{p, 2000} = \frac{100\text{ bar}}{1\text{ bar}} = 100

The standard Gibbs energy change at 2000 K2000\text{ K} is given by: ΔG∘(2000 K)=−RT2ln⁡(Kp,2000)=−R(2000)ln⁡(100)=−R(2000)⋅2ln⁡(10)=−4000Rln⁡(10)\Delta G^\circ (2000\text{ K}) = -R T_2 \ln(K_{p, 2000}) = -R (2000) \ln(100) = -R (2000) \cdot 2\ln(10) = -4000 R \ln(10)

Step 3: Ratio of Standard Gibbs Energy Changes The ratio of the standard Gibbs energy change at 1000 K1000\text{ K} to that at 2000 K2000\text{ K} is: ΔG∘(1000 K)ΔG∘(2000 K)=−1000Rln⁡(10)−4000Rln⁡(10)=10004000=0.25\frac{\Delta G^\circ (1000\text{ K})}{\Delta G^\circ (2000\text{ K})} = \frac{-1000 R \ln(10)}{-4000 R \ln(10)} = \frac{1000}{4000} = 0.25

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