Consider the equilibrium below: 2NOBr (g) doubleheadarrow 2NO (g) + Br_2 (g) Thi
ID: 1061279 • Letter: C
Question
Consider the equilibrium below: 2NOBr (g) doubleheadarrow 2NO (g) + Br_2 (g) This reaction was allowed to come to equilibrium at 25 degree C, and it was found that 34% of the NOBr had dissociated. If the total pressure was 0.25 atm. What is K_p for the reaction at this temperature? The equilibrium constant K_c is 1.2 at 375 degree C for the reaction below. If you started with [H_2] of 0.70M, 0.60M, and [NH_3] of 0.48M, which gases will increase in concentration and which will decrease is concentration as the reaction comes to equilibrium? N_2 (g) + 3H_2 (g) doubleheadarrow 2NH_3(g)Explanation / Answer
(1)
2 NOBr (g) <---------> 2 NO (g) + Br2 (g)
Initial : 1 0 0
Change : -2x +2x +x
Equilibrium: 1-2x +2x +x
Total moles = 1+x
Given that,
1-2x = 0.66
x= 0.17
Kp = ( (2x/1+x)*(P))2 x (x/1+x)*(P) / (( (1-2x) /(1+x) ) *P )2
Put x= 0.17 and P =0.25
Kp = 0.00636