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?

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)

To determine the equilibrium constant, Kc, for the second reaction, you cannot simply divide the equilibrium constant of the first reaction by 2. The equilibrium constant depends on the stoichiometry of the reaction, specifically the coefficients of the reactants and products.

To solve this problem, you need to use the concept of equilibrium constants and the equation relating equilibrium constants to stoichiometric coefficients. The equilibrium constant expression for the first reaction is:

Kc = [NO]^4[H2O]^6/[NH3]^4[O2]^5

Now, let's write the balanced equation for the second reaction:

2 NH3(g) + 5/2 O2(g) ⇌ 2 NO(g) + 3 H2O(g)

Note that the stoichiometric coefficients in the second reaction are different from the first reaction. In order to determine the equilibrium constant for the second reaction, we need to use the equation relating equilibrium constants to stoichiometric coefficients:

Kc2 = (Kc1)^n

Where Kc1 is the equilibrium constant for the first reaction, and n is the sum of the stoichiometric coefficients of the products in the second reaction minus the sum of the stoichiometric coefficients of the reactants in the second reaction:

n = (2 + 3) - (2 + 5/2)

n = 2 - 1/2

n = 1.5

Now, substitute the values into the equation:

Kc2 = (0.0015)^(1.5)

Using a calculator, you can calculate the value of Kc2:

Kc2 ≈ 0.0036

Therefore, the equilibrium constant, Kc, for the second reaction is approximately 0.0036.