If Kc = 7.5E-9 at 1000K for the reaction N2(g)+ O2(g)<=>2NO(g), what is Kc at 1000K for the reaction 2NO(g)<=>N2(g)+ O2(g)?
1.3×10^8
A + B ==> C, Keq = 10, then
C ==> A + B, Keq = 1/10
To find the value of Kc at 1000 K for the reaction 2NO(g) ⇌ N2(g) + O2(g), we can use the relationship between the equilibrium constants of the forward and reverse reactions.
For a reaction aA + bB ⇌ cC + dD, the equilibrium constant (Kc) of the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. Mathematically, we can express this as:
Kc (reverse) = 1 / Kc (forward)
In the given reaction:
N2(g) + O2(g) ⇌ 2NO(g)
The equilibrium constant (Kc) is given as 7.5E-9 at 1000 K.
Now, to find the equilibrium constant for the reverse reaction 2NO(g) ⇌ N2(g) + O2(g), we can use the relationship mentioned earlier:
Kc (reverse) = 1 / Kc (forward)
Substituting the given value of Kc (forward), we get:
Kc (reverse) = 1 / (7.5E-9)
To calculate the value of 1 / (7.5E-9), we divide 1 by 7.5E-9:
Kc (reverse) ≈ 1.333E8
Therefore, the value of Kc at 1000 K for the reaction 2NO(g) ⇌ N2(g) + O2(g) is approximately 1.333E8.