If the reaction quotient Q has a greater value than the related equilibrium cons
ID: 950394 • Letter: I
Question
If the reaction quotient Q has a greater value than the related equilibrium constant, K the reaction is at equilibrium the reaction will release will release heat to achieve equilibrium the reaction will continue to make more products the reaction will consume products and make reactants the value of K will increase until it is equal to Q In the equilibrium expressions. the concentration of pure solids and liquids have the assigned value of zero are treated as any other solute have constant values, c_3 and c_2 mass are determined from density and molar have the assigned value of one In the reaction: 2A 3B Delta G degree_formation of B is 215 KJ / mol Determine the equilibrium constant K at 100k (K = e^-Delta G degree / RT and R = 8.314 J/mol K) For the reaction: 2NO(g) + O_2(g) rightarrow 2NO_2(g) Determine the rate law: rate = K[NO]^m [O_2]^n Determine also the rate constant: K = (rate)/[NO]^m [O_2]^n rate = k[NO]^2 [O]^1 K = 1.77 K = [products]^p / [reactants]^T K = e^-Delta G degree / RT R = 8.314 J/mol*K Delta G degree rxn = [coefficient 8 Delta G degree formation] products-[coefficient * Delta G degree formation]reactantsExplanation / Answer
1)
K = e-delta G / RT
K = e -(215 *1000 J mol-1 / ( 8.314 J mol-1 K-1 * 1000 K ))
K = 5.877 * 10-12
2) a)
Rate = k [NO]m [O2]n
For experiment 1
0.048 = k ( 0.30 )m (0.30)n --------------- (1)
For experiment 2
0.192 = k ( 0.60 )m (0.30)n --------------- (2)
(2) / (1) ;
4 = (2) m
m = 2 (order of NO with respect to O2 )
For experiment 2
0.192 = k ( 0.60 )m (0.30)n --------------- (2)
For experiment 3
0.768 = k ( 0.60 )m (1.20)n --------------- (3)
(3) / (2) ;
4 = (4)n
n = 1 (order of O2 with respect to NO )
Total order of the reaction = 2 +1 = 3
So therefore the rate law of the reaction, Rate = k [NO]2 [O2]
b)
Rate = k [NO]2 [O2]
For experiment 1,
Rate = k (0.3 M )2 (0.3 M) = 0.048 M s-1
k = 1.78 M-2 s-1