Question

two moles of NH3 gas are introduced into a previously evacuated one litre vessel in which it partially dissociate at high temperature as 2NH3-N2+3H2.at equilibrium one mole of NH3 Remain.the value of Kc is??

Answer 1

Answer: $k_c$  for this reaction is 1.6875.

Explanation: We are given,

Volume of the vessel = 1L

For a reaction,

            $2NH_3\rightleftharpoons N_2+3H_2$

at t = 0      2            0         0

at  $t=t_{eq}$    2-2x        x       3x

at t = t        0            x         3x

where, 2x = change in moles of $NH_3$

It is given that at equilibrium, amount of $NH_3$ left is 1 mole,

$2-2x=1\\2x=1\\x=0.5$

Concentration can be calculated as

$C=\frac{m}{V}$

So, the concentraction of $NH_3$, $H_2$ and $N_2$ at equilibrium will be,

$[NH_3]=\frac{1}{1}$ = 1.0 mol/L

$[N_2]=\frac{0.5}{1}$ = 0.5 mol/L

$[H_2]=\frac{3\times 0.5}{1}$ = 1.5 mol/L

For a given reaction, $k_c$ can be calculated as:

$k_c=\frac{[N_2]^1[H_2]^3}{[NH_3]^1}\\k_c=\frac{(0.5)^1\text{mol/L }\times (1.5)^3\text{mol/L }}{(1)^1mol/L}$

$k_c$ = 1.6875

Answer 2

Answer:

the right answer of this question is 6.75

Explanation:

hope my answer will help you a lot