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2 years ago

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At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.200 M and [ NO ] = 0.400 M . N 2 ( g )

+ O 2 ( g ) − ⇀ ↽ − 2 NO ( g ) If more NO is added, bringing its concentration to 0.700 M, what will the final concentration of NO be after equilibrium is re‑established?

1 answer:

stiks02 [169]2 years ago

4 0

<u>Answer:</u> The equilibrium concentration of NO after it is re-established is 0.55 M

<u>Explanation:</u>

For the given chemical equation:

$N_2(g)+O_(g)\rightleftharpoons 2NO(g)$

The expression of $K_{eq}$ for above equation follows:

$K_{eq}=\frac{[NO]^2}{[N_2][O_2]}$ .....(1)

We are given:

$[NO]_{eq}=0.400M$

$[N_2]_{eq}=0.200M$

$[O_2]_{eq}=0.200M$

Putting values in expression 1, we get:

$K_{eq}=\frac{(0.400)^2}{0.200\times 0.200}\\\\K_{eq}=4$

Now, the concentration of NO is added and is made to 0.700 M

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

The equilibrium will shift in backward direction.

$N_2(g)+O_(g)\rightleftharpoons 2NO(g)$

<u>Initial:</u> 0.200 0.200 0.700

<u>At eqllm:</u> 0.200+x 0.200+x 0.700-2x

Putting values in expression 1, we get:

$4=\frac{(0.700-2x)^2}{(0.200+x)\times (0.200+x)}\\\\x=0.075$

So, equilibrium concentration of NO after it is re-established = (0.700 - 2x) = [0.700 - 2(0.075)] = 0.55 M

Hence, the equilibrium concentration of NO after it is re-established is 0.55 M

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Putting values in above expression, we get:

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To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

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