At 309 k, 309 k, this reaction has a k c kc value of 0.0504 . 0.0504. x ( g ) + 3 y ( g ) − ⇀ ↽ − 2 z ( g ) x(g)+3y(g)↽−−⇀2z(g) calculate k p kp at 309 k. 309 k. note that the pressure is in units of atmosphere (atm).

Kp = 7.85x10^-5

Explanation:

To do this, let's write again the given equation:

X + 3Y <> 2Z     Kc = 0.0504

We know that the expression for Kc would be:

Kc = [Z]² / [X] * [Y]³

And the expression for pressure is:

P = nRT/V

The expression of n/V is concentration so

P = CRT  

To calculate the value of Kp, is the same for Kc but instead of using concentrations, we use the pressure:

Kp = pZ² / pX * pY³

After we replace the values of p into the above equation, we have the following:

Kp = ([Z]RT)² / ([X]RT) * ([Y]RT)³

Kp = [Z]²(RT)² / [X]RT * [Y]³(RT)³ --->Here we take common factor of RT so

Kp = [Z]² / [X] * [Y]³ * (RT)^(2 - (1 + 3))

The part in bolding is the value of Kc so:

Kp = Kc * (RT)^-2

Replacing the values now of Kc, R (0.082 L atm/ K mol) and T, let's calculate Kp:

Kp = 0.0504 * (0.082 * 309)^-2

Kp = 7.85x10^-5

possibly, yes.

explanation:

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