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Byju's Answer

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 $2000 K$ ?

0.2500

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0.25

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.250

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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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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Similar questions

Q. Consider the reaction

A ⇌ B

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

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. Given below is the plot of the partial pressure of

B

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What is the ratio of the standard Gibbs energy of the reaction at

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Q. 13.8g of

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was placed in a 1 L reaction vessel at 400 K and allowed to attain equilibrium

N_{2} O_{4} ⇌ 2 N O_{2}

The total pressure at equilibrium was found to be 9.15bar. Calculate

K_{P}, K_{C}

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NO and $B r_{2}$ at initial partial pressure of 98.4 and 41.3 torr, respectively were allowed to react at 300K. At equilibrium the total pressure was 110.5 torr. Calculate the value of KP for the following reaction at 300K.

$2 N O (g) + B r_{2} (g) ⇌ 2 N O B r (g)$

13.8 g

N_{2} O_{4}

was placed in a

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and allowed to attain equilibrium

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

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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

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

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

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

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