The equilibrium constant for the reaction:
2NO(g)+Br2(g)<==>2NOBr(g)
Kc=1.3x10^-2 at 1000 Kelvin
Calculate Kc for:
NOBr(g)<==>NO(g)+ 1/2Br2(g)
8.8 ...... this is 100% the answer
Call Keq for equation 1 k1.
Call Keq for equation 2 k2.
k2 = sqrt(1/k1).
5.8
To calculate the equilibrium constant (Kc) for the reverse reaction, which is:
NOBr(g) <==> NO(g) + 1/2 Br2(g)
You can use the relationship between the forward and reverse reactions in terms of their equilibrium constants.
For a reaction in the form:
aA + bB <==> cC
The equilibrium constant (Kc) can be expressed as:
Kc = ([C]^c) / ([A]^a * [B]^b)
Where [A], [B], and [C] represent the molar concentrations of substances A, B, and C, respectively.
In this case, you need to find the equilibrium constant (Kc) for the reverse reaction, so you can express it in terms of the equilibrium constant (Kc') for the forward reaction:
NO(g) + 1/2 Br2(g) <==> NOBr(g)
Since the coefficients of the balanced equation for the forward reaction are halved when compared to the reverse reaction, Kc' can be written as:
Kc' = (Kc)^(-1/2)
Substituting the given value of Kc = 1.3 x 10^-2 into the equation gives:
Kc' = (1.3 x 10^-2)^(-1/2)
To simplify this expression, calculate the square root of (1.3 x 10^-2) and then calculate the reciprocal:
Kc' = 1 / √(1.3 x 10^-2)
Kc' ≈ 1 / 0.113
Kc' ≈ 8.85
Therefore, the equilibrium constant (Kc) for the reverse reaction:
NOBr(g) <==> NO(g) + 1/2 Br2(g)
is approximately 8.85.