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aleksandrvk [35]

2 years ago

12

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibrium described by the equation N2O4(g)

2NO2(g) If at equilibrium the N2O4 is 39% dissociated, what is the value of the equilibrium constant, Kc, for the reaction under these conditions

1 answer:

lora16 [44]2 years ago

4 0

<u>Answer:</u> The value of equilibrium constant is 0.997

<u>Explanation:</u>

We are given:

Percent degree of dissociation = 39 %

Degree of dissociation, $\alpha$ = 0.39

Concentration of $N_2O_4$ , c = $\frac{1mol}{1L}=1M$

The given chemical equation follows:

$N_2O_4\rightleftharpoons 2NO_2$

<u>Initial:</u> c -

<u>At Eqllm:</u> $c-c\alpha$ $2c\alpha$

So, equilibrium concentration of $N_2O_4=c-c\alpha =[1-(1\times 0.39)]=0.61M$

Equilibrium concentration of $NO_2=2c\alpha =[2\times 1\times 0.39]=0.78M$

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

$K_{c}=\frac{[NO_2]^2}{[N_2O_4]}$

Putting values in above equation, we get:

$K_{c}=\frac{(0.78)^2}{0.61}\\\\K_{c}=0.997$

Hence, the value of equilibrium constant is 0.997

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Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

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On addition of product at equilibrium shifts the equilibrium in backward direction.
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$A(aq)+B(aq)\rightleftharpoons 2C(aq)$

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1. Increase A

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On increasing the amount of B at equilibrium will shift the equilibrium in forward or rightward direction.

3. Increase C

On increasing the amount of C at equilibrium will shift the equilibrium in backward or leftward direction.

4. Decease A

On decreasing the amount of A at equilibrium will shift the equilibrium in backward or leftward direction.

5. Decease B

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On decreasing the amount of C at equilibrium will shift the equilibrium in forward or rightward direction.

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On doubling A and halving B, equilibrium constant of the reaction = K'

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The value of equilibrium constant K' is equal to K, which means that equilibrium will not shift in any direction.

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On doubling B and C, equilibrium constant of the reaction = K'

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20.0 g of Mg reacts with excess HCl. To calculate the no. of grams of HCl that reacted, we should calculate the no. of moles of Mg:

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