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Derivation of Kp and Kc

N2O2⇋ 2NO: K1...

Question

$N_{2} O_{2} ⇋ 2 N O : K_{1}; 1 / 2 N_{2} + 1 / 2 O_{2} ⇋ N O : K_{2}$
$2 N O ⇋ N_{2} + O_{2} : K_{3}; N O ⇋ 1 / 2 N_{2} + 1 / 2 O_{2} : K_{4}$
Correct relation between $K_{1}, K_{2}, K_{3}$ and $K_{4}$ is:

A

K_{1} \times K_{3} = 1

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B

\sqrt{K_{1}} \times K_{4} = 1

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C

\sqrt{K_{3}} \times K_{2} = 1

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D

None of these

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Solution

The correct option is A $\sqrt{K_{3}} \times K_{2} = 1$
$K_{1} \times K_{3} = \frac{[N O]^{2}}{[N_{2} O_{2}]} \times (\frac{[N_{2}] [O_{2}]}{[N O]^{2}}) = (\frac{[N_{2}] [O_{2}]}{[N_{2} O_{2}]}) \neq 1$
$\sqrt{K_{3}} \times K_{2} = \sqrt{(\frac{[N_{2}] [O_{2}]}{[N O]^{2}})} \times (\frac{[N O]}{[N_{2}]^{\frac{1}{2}} [O_{2}]^{\frac{1}{2}}}) = 1$

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Similar questions

N_{2} O_{2} ⇌ 2 N O, K_{1}; (\frac{1}{2}) N_{2} + (\frac{1}{2}) O_{2} ⇌ N O, K_{2}

;

2 N O ⇌ N_{2} + O_{2}, K_{3}; N O ⇌ (\frac{1}{2}) N_{2} + (\frac{1}{2}) O_{2}, K_{4}

;
Correct relation (s) between

K_{1}, K_{2}, K_{3}

K_{4}

N_{2} + O_{2} ⇌ 2 N O, K_{1}

;

(\frac{1}{2}) N_{2} + (\frac{1}{2}) O_{2} ⇌ N O, K_{2}

;

2 N O ⇌ N_{2} + O_{2}; K_{3}

;

N O ⇌ (\frac{1}{2}) N_{2} + (\frac{1}{2}) O_{2}; K_{4}

Correct relation(s) between

K_{1}, K_{2}, K_{3}

K_{4}

N_{2} + O_{2} ⇋ 2 N O, K_{1}; \frac{1}{2} N_{2} + \frac{1}{2} O_{2} ⇋ N O, K_{2};

2 N O ⇌ N_{2} + O_{2}, K_{3}; N O ⇌ \frac{1}{2} N_{2} + \frac{1}{2} O_{2}, K_{4}

Correct relation(s) between

K_{1}, K_{2}, K_{3}

K_{4}

For these chemical equilibriums:

N_{2} + O_{2} ⇋ 2 N O, K_{1}

\frac{1}{2} N_{2} + \frac{1}{2} O_{2} ⇋ N O, K_{2}

2 N O ⇌ N_{2} + O_{2}; K_{3}

N O ⇌ \frac{1}{2} N_{2} + \frac{1}{2} O_{2}, K_{4}

K_{1} \times K_{3} = x; \sqrt{(K_{1})} \times K_{4} = y; \sqrt{(K_{3})} \times K_{2} = z

What will be the value of

x y z

Q. Given four reactions and their equilibrium constants.

N_{2} O_{2} ⇌ 2 N O, K_{1}; \frac{1}{2} N_{2} + \frac{1}{2} O_{2} ⇌ N O, K_{2};

2 N O ⇌ N_{2} + O_{2}, K_{3}; N O ⇌ \frac{1}{2} N_{2} + \frac{1}{2} O_{2}, K_{4}

Correct relation between

K_{1}, K_{2}, K_{3} a n d K_{4}

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