Given the following equilibria: K_1 : N_2 + 3 | vedclass

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

Question

(C) The given reactions are: $(1) N_2 + 3H_2 \rightleftharpoons 2NH_3$ with constant $K_1$ $(2) N_2 + O_2 \rightleftharpoons 2NO$ with constant $K_2$

Vedclass · Accepted Answer

$\frac{K_2 K_3^3}{K_1}$

Vedclass · Answer

(C) The given reactions are: $(1) N_2 + 3H_2 ightleftharpoons 2NH_3$ with constant $K_1$ $(2) N_2 + O_2 ightleftharpoons 2NO$ with constant $K_2$ $(3) H_2 + 1/2 O_2 ightleftharpoons H_2O$ with constant $K_3$ We want to obtain the reaction: $2NH_3 + 5/2 O_2 ightleftharpoons 2NO + 3H_2O$ To get this,we reverse reaction $(1)$,add reaction $(2)$,and add $3$ times reaction $(3)$: Reverse $(1): 2NH_3 ightleftharpoons N_2 + 3H_2$ (Constant $= 1/K_1$) $(2): N_2 + O_2 ightleftharpoons 2NO$ (Constant $= K_2$) $3 	imes (3): 3H_2 + 3/2 O_2 ightleftharpoons 3H_2O$ (Constant $= K_3^3$) Adding these: $(2NH_3 + N_2 + 3H_2 + O_2 + 3/2 O_2) ightleftharpoons (N_2 + 3H_2 + 2NO + 3H_2O)$ Simplifying: $2NH_3 + 5/2 O_2 ightleftharpoons 2NO + 3H_2O$ The equilibrium constant $K$ is: $K = (1/K_1) 	imes K_2 	imes K_3^3 = \frac{K_2 K_3^3}{K_1}$.

Given the following equilibria:
$K_1 : N_2 + 3H_2 \rightleftharpoons 2NH_3$
$K_2 : N_2 + O_2 \rightleftharpoons 2NO$
$K_3 : H_2 + 1/2 O_2 \rightleftharpoons H_2O$
The equilibrium constant for the reaction $2NH_3 + 5/2 O_2 \rightleftharpoons 2NO + 3H_2O$ in terms of $K_1, K_2$,and $K_3$ is:

Similar Questions

The equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is $32$ at a given temperature. The equilibrium concentrations of $I_2$ and $HI$ are $0.5 \times 10^{-3} \ M$ and $8 \times 10^{-3} \ M$ respectively. The equilibrium concentration of $H_2$ is:

The value of $K_{c}$ for the reaction $3 O_{2(g)} \longleftrightarrow 2 O_{3(g)}$ is $2.0 \times 10^{-50}$ at $25^{\circ} C$. If the equilibrium concentration of $O_{2}$ in air at $25^{\circ} C$ is $1.6 \times 10^{-2} \, M$,what is the concentration of $O_{3} ?$

Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

If the equilibrium constant for the reaction,$H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$ is $K$,what is the equilibrium constant of $HI_{(g)} \rightleftharpoons \frac{1}{2} H_{2(g)} + \frac{1}{2} I_{2(g)}$?

The decomposition of $N_2O_4$ to $NO_2$ is carried out at $280 \ K$ in chloroform. When equilibrium has been established,$0.2 \ mol$ of $N_2O_4$ and $2 \times 10^{-3} \ mol$ of $NO_2$ are present in $2 \ L$ solution. The equilibrium constant for the reaction $N_2O_4 \rightleftharpoons 2NO_2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Given the following equilibria: $K_1 : N_2 + 3H_2 \rightleftharpoons 2NH_3$ $K_2 : N_2 + O_2 \rightleftharpoons 2NO$ $K_3 : H_2 + 1/2 O_2 \rightleftharpoons H_2O$ The equilibrium constant for the reaction $2NH_3 + 5/2 O_2 \rightleftharpoons 2NO + 3H_2O$ in terms of $K_1, K_2$,and $K_3$ is: