In which of the following reactions is the va | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$.For $K_p$ to be equal to $K_c$,the term $(RT)^{\De

Vedclass · Accepted Answer

$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$

Vedclass · Answer

(A) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$.For $K_p$ to be equal to $K_c$,the term $(RT)^{\Delta n_g}$ must be equal to $1$,which occurs when $\Delta n_g = 0$.$\Delta n_g$ is defined as the difference between the sum of stoichiometric coefficients of gaseous products and gaseous reactants: $\Delta n_g = \sum n_p - \sum n_r$.For option $A$: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$,$\Delta n_g = 2 - (1 + 1) = 0$.Since $\Delta n_g = 0$,$K_p = K_c$ for this reaction.

In which of the following reactions is the value of $K_p$ equal to $K_c$?

Similar Questions

$K_p$ for the reaction $2SO_3 \rightleftharpoons 2SO_2 + O_2$ at $700 \ K$ is $1.3 \times 10^{-3} \ atm$. The $K_c$ at the same temperature for the reaction $2SO_2 + O_2 \rightleftharpoons 2SO_3$ will be:

For the formation of ammonia gas from its constituent elements,the $K_{P} / K_{C}$ is

In which of the following equilibria,the value of $K_p$ is less than $K_c$?

The equilibrium constant for the reversible reaction,$N_2 + 3H_2 \rightleftharpoons 2NH_3$ is $K$ and for the reaction $\frac{1}{2}N_2 + \frac{3}{2}H_2 \rightleftharpoons NH_3$ the equilibrium constant is $K'$. $K$ and $K'$ will be related as

The equilibrium $NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$ is set up at $127 \,^oC$ in a closed vessel. The total pressure at equilibrium was $20 \,atm$. The $K_C$ for the reaction is: (in $,M^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module