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Vant Hoff's Equation and Effect of Temperature on Equilibrium Constant

For the react...

Question

For the reaction $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ , the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is correct?

The reaction is endothermic.

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The reaction is exothermic.

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N O_{2} (g)

and

N_{2} O_{4} (g)

are mixed at 400 K at partial pressures 20 bar and 2 bar respectively, more

N_{2} O_{4} (g)

will be formed.

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The entropy of the system increases.

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Solution

The correct options are
A The reaction is endothermic.
C If $N O_{2} (g)$ and $N_{2} O_{4} (g)$ are mixed at 400 K at partial pressures 20 bar and 2 bar respectively, more $N_{2} O_{4} (g)$ will be formed.
D The entropy of the system increases.
As the temperature increases from 400 K to 500 K, the value of K increases from 50 to 1700.
Thus with increase in temperature, the equilibrium shifts to right.
This is possible for an endothermic reaction because, the heat absorbed will nullify the effect of increased temperature.
Hence, the forward reaction is endothermic.
The expression for the reaction quotient is $Q = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]}$ .
Substitute values in the above expression.
$Q = \frac{[20]^{2}}{[2]} = 200$
It is greater than the value of equilibrium constant which is 50. Thus more of reactant $(N_{2} O_{4})$ will be formed.
During the reaction, the number of moles increases from reactants to products. Hence, the entropy of the system increases.

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Similar questions

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For the reaction N2O4(g)⇌2NO2(g), the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is correct?

For the reaction $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ , the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is correct?