N2+O2 <-> 2NO

Equillibrium Concentrations of the gases at 1500K are 1.7x10(-3) mol/L for o2 and 6.4x10(-3) mol/L for N2 and 1.1x10(-5)mol/L for NO. CAlculate the value of Kc at 1500K for these data?

Show work please

Kc = (NO)^2/(N2)(O2)

Substitute the values from the problem into the Kc expression and solve for Kc. It's that simple.

To calculate the value of Kc at 1500K for this equilibrium reaction, we need to use the equation:

Kc = ([NO]²) / ([N₂] x [O₂])

Given that the equilibrium concentrations are 1.7x10^(-3) mol/L for O₂, 6.4x10^(-3) mol/L for N₂, and 1.1x10^(-5) mol/L for NO, we can substitute these values into the equation:

Kc = ([1.1x10^(-5)]²) / ([6.4x10^(-3)] x [1.7x10^(-3)])

First, let's square the concentration of NO:

(NO)² = (1.1x10^(-5))²
= 1.21x10^(-10)

Next, multiply the concentrations of N₂ and O₂:

([N₂] x [O₂]) = (6.4x10^(-3)) x (1.7x10^(-3))
= 1.088x10^(-5)

Now, substitute these values into the Kc equation:

Kc = (1.21x10^(-10)) / (1.088x10^(-5))

To simplify this expression, we divide the numerator by the denominator:

Kc = 1.11x10^(-10)

Therefore, the value of Kc at 1500K for these data is approximately 1.11x10^(-10).