Chemical EquilibriumHard
Question
Consider the following equilibrium in a closed container. N2O4(g) $\rightleftharpoons$2NO2(g)
At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statement holds true regarding the equilibrium constant (KP) and degree of dissociation (α)?
Options
A.Neither KP nor α changes.
B.Both KP and α changes.
C.KP changes, but α does not change.
D.KP does not change, but α changes.
Solution
Kp is a function of temperature only. On reducing the volume, equilibrium will shift backward and hence, α will decrease.
Create a free account to view solution
View Solution FreeMore Chemical Equilibrium Questions
For the following mechanism, P + Q PQ R at equilibrium is :[k represents rate constant]...The Haber’s process for the manufacture of ammonia is usually carried out at about 500oC. If a temperature of about 250o...PCl5(g) $\rightleftharpoons$PCl3(g) + Cl2(g). In the above reaction, the partial pressure of PCl3, Cl2 and PCl5 are 0.3,...One mole of ethanol is treated with one mole of ethanoic acid at 25oC. One-fourth of the acid changes into ester at equi...A certain weak acid has Ka = 1 × 10-4 . Calculate the equilibrium constant for its rx with a strong base :...