In the two gaseous reactions (i) and (ii) at `250^@C` (i) `2NO(g)+(1)/(2)O_2(g)hArrNO_2(g)K_1` (ii) `2NO_2(g) hArr2NO(g)+O_2(g),K_2` the equilibrium c

Question

In the two gaseous reactions (i) and (ii) at `250^@C` (i) `2NO(g)+(1)/(2)O_2(g)hArrNO_2(g)K_1` (ii) `2NO_2(g) hArr2NO(g)+O_2(g),K_2` the equilibrium c

In the two gaseous reactions (i) and (ii) at `250^@C`
(i) `2NO(g)+(1)/(2)O_2(g)hArrNO_2(g)K_1`
(ii) `2NO_2(g) hArr2NO(g)+O_2(g),K_2` the equilibrium constants `K_1` and `K_2` are releated as
A. `K_2=(1)/(K_1)`
B. `K_2=K_1^(1//2)`
C. `K_2=(K_1)/(K_1^2)`
D. `K_2=K_1^(2)`

Monjur Alahi

6 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - C
For equation (i),
`NO(g)(1)/(2)O_2(g)hArr NO_2(g)`
`K_1=([NO_2])/([NO][O_2]^(1//2))` ....(i)
For reaction (ii)
`2NO_2(g)hArr2NO(g)+O_2(g)K_2=([NO]^2[O_2])/([NO_2]^2)`.....(ii)
Now, on reversing equation (i), we get,
`(1)/(k_(1))=(1)/(([NO_(2)])/([NO][O_(2)]^(1//2)))=([NO][O_2]^(1//2))/([NO_2]^2)`
`(1/(K_1))^2={([NO][O_2]^(1//2))/([NO_2])}^2=([NO]^2[O_2])/([NO_2]^2)=K_2`