Consider the following reaction in a 1 L closed vessel.
`N_2 + 3H_2 iff NH_3`
If all the species: `N_2, H_2` and `NH_3` are in 1 mol in the beginning of the reaction and equilibrium is attained after unreacted `N_2` is 0.7 mol. What is the value of equilibrium constant?

Options

A. 3600
B. 3657.14
C. 2657.14
D. 1828.57

Answer

B

Explanation

For the given reaction :
` (##ARH_5Y_SP_05_05_18_01_E03_012_S01.png" width="80%">
given , 1 - x = 0.7mol
` therefore x = 0.3 ` mol
therefore , concentration of `N_2, H_2 ` and `NH_3` at equilibrium will be
` [N_2] = [ 0.7]`
`[H_2] = 1- (3 xx 0.3) = [0.1]`
` [NH_3] = 1 + 2x = 1 + (2 xx 0.3) = [1.6]`
According to law of equilibrium constant `[K_c]`
`K_c = ([NH_3]^2)/([N^2] , [H_2]^3) = ([1.6]^2)/([0.7][0.1]^3)`
`K_c = ((2.56) )/(0.007) = 3557.14`
Hence . equilirbrium constant `(K_c)` = 3657.14 `