2 years ago

Please help me on this.

Chemistry

1 answer:

Anestetic [448]2 years ago

3 0

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Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature,

7nadin3 [17]

Answer:

Kc = 1.54e - 31 / 2.61e - 24

Explanation:

1 ) $N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)$ ; Kc = 1.54e - 31

2) $N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)$ ; Kc = 2.16e - 24

upon reversing ( 2 ) equation

$N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)$ Kc = 1/2.16e - 24

now adding 1 and reversed equation (2)

$N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)$

$N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)$

we get ,

$N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)$ Kc = 1.54e-31 × 1/2.61e - 24

equilibrium constant of equation (3) is -

Kc = 1.54e - 31 / 2.61e - 24

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