nope.
let the concentrations of NH3, O2, NO, H2O be A,B,C,D
Kc=C^2 * D^3/A^2B^(5/2)
and
.0015=C^4D^6/A^4B^6
but the second right hand term is the square of the first
kc^2=.0015
Kc=sqrt(.0015)
Rxn 1: 4 NH3(g) + 5 O2(g) ⇌ 4 NO(g) + 6 H2O(g)
If the equilibrium constant, Kc, for Rxn 1 is 0.0015, determine the equilibrium constant, Kc, for the
reaction: 2 NH3(g) + 5/2 O2(g) ⇌ 2 NO(g) + 3 H2O(g)
I'm not sure how to start this problem would it possibly just be half (7.5x10-4) since the second reaction is just half of the first?
1 answer