2 years ago

This has been deleted as question wasn't answered on time

Chemistry

1 answer:

astra-53 [7]2 years ago

8 0

How i be feeling waiting for my questions to be answered in the middle of a quiz.

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The equilibrium constant, Kc, for the following reaction is 0.967 at 650 K. 2NH3(g) N2(g) 3H2(g) When a sufficiently large sampl

AlekseyPX

Answer: Concentration of $NH_3$ in the equilibrium mixture is 0.31 M

Explanation:

Equilibrium concentration of $H_2$ = 0.729 M

The given balanced equilibrium reaction is,

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

Initial conc. x 0 0

At eqm. conc. (x-2y) M (y) M (3y) M

The expression for equilibrium constant for this reaction will be:

3y = 0.729 M

y = 0.243 M

$K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}$

Now put all the given values in this expression, we get :

$K_c=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}$

$0.967=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}$

$x=0.80$

concentration of $NH_3$ in the equilibrium mixture = $0.80-2\times 0.243=0.31$

Thus concentration of $NH_3$ in the equilibrium mixture is 0.31 M

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