Question

Given the following: I) N20(g) 1/2 02(g) 2 NO(g) II) N2(g) 02(g) 2 NO(g) Ke= 4.1 x 10-31 Ke= 1.7 x 10-13 Find the value of the equilibrium constant for the following equilibrium reaction: N2(g) 1/202(g) N20(g) A) 7.0 x 10-44 B) 4.2 x 1017 C) 2.4 x 10-18 D) 1.6 x 109 E) 2.6 x 10-22

Answer 1

Answer:

Answer B) 4.2x10^17

Explanation:

To produce the reaction 3 using reaction 1 and 2 we need to invert the order of the first reaction the second in the same order, as it's shown:

$2NO-------->N_2O+1/2O_2 : (reaction 1' :K'_1 =1/K_1=2.4x10^ 3^0)$

$N_2+O_2-------->2NO: (reaction 2 :K_2 =1.7x10^-^1^3)$

____________________________

$N_2+1/2O_2-------->N_2O: (reaction 3)$

Due to the inversion of the first equation, the equilibrium constant of the new reaction is K1'=1/K1.=2.4x10^30

Finally, the new equilibrium constant K3 is the product of the previous constants:  

K3=K1'*K2=4.2x10^17