Two moles of NH_3 when put into a previously | vedclass

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

Question

(D) The balanced chemical equation is: $2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$ Initially,$2 \ mol$ of $NH_3$ are present in a $1 \ L$ vessel. A

Vedclass · Accepted Answer

$27/16 \ mol^2 \ L^{-2}$

Vedclass · Answer

(D) The balanced chemical equation is: $2NH_3(g) ightleftharpoons N_2(g) + 3H_2(g)$ Initially,$2 \ mol$ of $NH_3$ are present in a $1 \ L$ vessel. At equilibrium,$1 \ mol$ of $NH_3$ remains,meaning $1 \ mol$ has dissociated. According to the stoichiometry,$2 \ mol$ of $NH_3$ produce $1 \ mol$ of $N_2$ and $3 \ mol$ of $H_2$. Therefore,$1 \ mol$ of $NH_3$ dissociation produces $0.5 \ mol$ of $N_2$ and $1.5 \ mol$ of $H_2$. The equilibrium concentrations are: $[NH_3] = 1 \ mol/L$,$[N_2] = 0.5 \ mol/L$,and $[H_2] = 1.5 \ mol/L$. The equilibrium constant $K_c$ is given by: $K_c = \frac{[N_2][H_2]^3}{[NH_3]^2} = \frac{0.5 	imes (1.5)^3}{1^2} = 0.5 	imes 3.375 = 1.6875 = \frac{27}{16} \ mol^2 \ L^{-2}$.

Two moles of $NH_3$ when put into a previously evacuated vessel $(1 \ L)$,partially dissociate into $N_2$ and $H_2$. If at equilibrium one mole of $NH_3$ is present,the equilibrium constant is

Similar Questions

For the chemical equilibrium $A + B \rightleftharpoons C + D$,when one mole of each reactant is mixed,$0.4 \ mol$ of each product is formed. The equilibrium constant $K_c$ is:

Before any reaction takes place,the value of the reaction quotient $(Q_c)$ is ......

Explain the reaction quotient and how it is used to predict the direction of a reaction.

$PCl_5$,$PCl_3$,and $Cl_2$ are at equilibrium at $500 \ K$ with concentrations $[PCl_3] = 1.59 \ M$,$[Cl_2] = 1.59 \ M$,and $[PCl_5] = 1.41 \ M$. Calculate $K_c$ for the reaction:
$PCl_5 \rightleftharpoons PCl_3 + Cl_2$

For the reaction $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$,$K_c = 1.8 \times 10^{-6}$ at $185\,^{\circ}C$. At $185\,^{\circ}C$,the value of $K_c$ for the reaction $NO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons NO_{2(g)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module