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Given N(2)(g)+3H(2)(g)hArr2NH(3)(g),K(...

Given
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g),K_(1)`
`N_(2)(g)+O_(2)(g)hArr2NO(g),K_(2)`
`H_(2)(g)+(1)/(2)O_(2)hArrH_(2)O(g),K_(3)`
The equilibrium constant for
`2NH_(3)(g)+(5)/(2)O_(2)(g)hArr2NO(g)+3H_(2)O(g)`
will be

A

`K_(2)K_(3)^(3)//K_(1)`

B

`K_(2)K_(3)//K_(1)`

C

`K_(2)^(3)K_(3)//K_(1)`

D

`K_(1)K_(3)^(3)//K_(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`N_(2) + 3H_(2)hArr2NH_(3)" "K_(1) rarr` (1)
`N_(2) + O_(2)hArr2NO" "K_(2)rarr` (2)
`H_(2) + (1)/(2)O_(2)rarrH_(2)O" "K_(3)rarr` (3)
Now, equation (2) + 3 x equation (3) - equation (1)
`2NH_(3) + (5)/(2)O_(2)hArr 2NO + 3H_(2)O`
`implies K =K_(2.K_(3)^(3))/(K_(1))` , Option (1)
In equilibrium reaction when equations are added 'k' are multiplied and if equations are subracted 'K' is divided.
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Knowledge Check

  • The following equlibrium given by N_(2)+3H_(2)hArrZ_((g))2NH_(3),K_(1) N_(2)+O_(2)hArrH_(2)O,K_(2) H_(2)+1/2O_(2)hArrH_(2)O,K_(3) the equlibrium constant of the reaction 2NH_(3)+5/2O_(2)hArr2NO+3H_(2)O"in terms of"K_(1),K_(2)and K_(3) is

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  • If the equilibrium constant for N_(2)(g)+O_(2)(g)hArr2NO(g) is K, the equilibrium constant for 1/2N_(2)(g)+1/2O_(2)(g)hArr NO(g) will be

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