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maxonik [38]
3 years ago
14

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

e of NH3(g) is introduced into an evacuated vessel at 650 K, the equilibrium concentration of H2(g) is found to be 0.729 M. Calculate the concentration of NH3 in the equilibrium mixture. M
Chemistry
1 answer:
AlekseyPX3 years ago
3 0

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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Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:

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Regards!

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