28 , g of N_2 and 6 , g of H_2 were kept at 4 | vedclass

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

Question

(A) The balanced chemical equation is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ Initial moles: $n(N_2) = \frac{28 \, g}{28 \, g/mol} = 1 \, mol$

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(A) The balanced chemical equation is: $N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$ Initial moles: $n(N_2) = \frac{28 \, g}{28 \, g/mol} = 1 \, mol$,$n(H_2) = \frac{6 \, g}{2 \, g/mol} = 3 \, mol$,$n(NH_3) = 0 \, mol$ Equilibrium moles of $NH_3 = \frac{27.54 \, g}{17 \, g/mol} = 1.62 \, mol$ According to stoichiometry,$2 \, mol$ of $NH_3$ are formed from $1 \, mol$ of $N_2$ and $3 \, mol$ of $H_2$. So,$1.62 \, mol$ of $NH_3$ are formed from $0.81 \, mol$ of $N_2$ and $2.43 \, mol$ of $H_2$. Equilibrium moles: $n(N_2) = 1 - 0.81 = 0.19 \, mol$,$n(H_2) = 3 - 2.43 = 0.57 \, mol$,$n(NH_3) = 1.62 \, mol$ Since the volume is $1 \, L$,concentration $[X] = n(X) \, mol/L$. $K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(1.62)^2}{(0.19)(0.57)^3} \approx \frac{2.6244}{0.19 	imes 0.185} \approx \frac{2.6244}{0.03515} \approx 74.66 \approx 75$.

$28 \, g$ of $N_2$ and $6 \, g$ of $H_2$ were kept at $400 \, ^oC$ in a $1 \, L$ vessel. The equilibrium mixture contained $27.54 \, g$ of $NH_3$. The approximate value of $K_c$ for the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ is (in $L^2 \, mol^{-2}$):

Similar Questions

The equilibrium constant at $298 \ K$ for a reaction $A + B \rightleftharpoons C + D$ is $100$. If the initial concentration of all the four species were $1 \ M$ each,then the equilibrium concentration of $D$ (in $mol \ L^{-1}$) will be:

At $717 \ K$,$3.2 \ mol$ of $HI$ is heated in a closed tube. $20\%$ of $HI$ decomposes at equilibrium according to the reaction $2HI_{(g)} \rightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$. Find $K_c$ and the moles of $HI$,$H_2$,and $I_2$ at equilibrium.

For the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,the forward reaction at constant temperature is favoured by

When the reaction $A + 2B \rightleftharpoons 2C + D$ was studied,it was observed that the initial concentration of $B$ was $1.5$ times that of $A$,and the equilibrium concentrations of $A$ and $C$ were equal. Then $K_C$ for the given equilibrium is equal to

$A$ plot of $\ln K$ against $\frac{1}{T}$ ($x$-axis) is expected to be a straight line,with intercept on $Y$-axis equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module