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Standard XII

Chemistry

Visualising Orbitals

Consider the ...

Question

Consider the reaction $A ⇌ B$ at $1000 K$ . At time $t$ , the temperature of the system was increased to $2000 K$ and the system was allowed to reach equilibrium. Throughout this experiment, the partial pressure of $A$ was maintained at $1 b a r$ . Given below is the plot of the partial pressure of $B$ with time.
What is the ratio of the standard Gibbs energy of the reaction at $1000 K$ to that at $2000 K$ ?

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Solution

$K_{e q} = \frac{[P_{B}]}{[P_{A}]}; {[P_{A}] = 1 atm}$

$K_{e q .2000 K} = 100$

$K_{e q .1000 K} = 10$

$\frac{Δ G_{1000 K}^{\circ}}{Δ G_{2000 K}^{\circ}} = \frac{{(- R T l n K_{e q})}_{1000 K}}{{(- R T ln K_{e q})}_{2000 K}} = \frac{1000 \times l n 10}{2000 \times l n 100} = \frac{1}{4} = 0.25$

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Keq=[PB][PA] ; {[PA]=1 atm} Keq.2000 K=100 Keq.1000 K=10 ΔG∘1000 KΔG∘2000 K=(−RT lnKeq)1000 K(−RTlnKeq)2000K=1000×ln102000×ln100=14=0.25

$K_{e q} = \frac{[P_{B}]}{[P_{A}]}; {[P_{A}] = 1 atm}$

$K_{e q .2000 K} = 100$

$K_{e q .1000 K} = 10$

$\frac{Δ G_{1000 K}^{\circ}}{Δ G_{2000 K}^{\circ}} = \frac{{(- R T l n K_{e q})}_{1000 K}}{{(- R T ln K_{e q})}_{2000 K}} = \frac{1000 \times l n 10}{2000 \times l n 100} = \frac{1}{4} = 0.25$