One wacky chemist wanted to get weird and study the relationship between A and D

Question

One wacky chemist wanted to get weird and study the relationship between A and D3E in the following equilibrium:
A + B + 3D + 2/3E CD + 2/3D3E
What is the concentration-based equilibrium constant, Kc, for this equilibrium at 26.91°C?

This is the info from previous questions

A + B C

[A] = 0.134 M
[B] = 0.320 M
[C] = 0.516 M

_

C + 3D CD + D2
PC = 0.0153 atm
PD = 0.0754 atm
PCD = 0.764 atm
PD2 = 0.155 atm

_

3D2 + 2E 2D3E
[D2] = 0.529 M
[E] = 0.0602 M
[D3E] = 0.000896 M

I was able to solve this (15100000)

What is the pressure-based equilibrium constant, Kp, for this equilibrium at the same temperature?

Should be 41.0 but I don't know how to get that.

Explanation / Answer

Solution:- Kp = Kc(RT)delta n

Where, delta n = sum of coefficients of products - sum of coefficients of reactants

The over all balanced equation is given as...

A + B + 3D + 2/3E CD + 2/3D3E

delta n = (1+2/3) - (1+1+3+2/3)

delta n = -4

Kc = 15100000

R = 0.0821

T = 26.91 + 273 = 299.91

Kp = 15100000(0.0821 x 299.91)-4

Kp = 15100000(24.62)-4

Kp = 15100000(0.00000272)

Kp = 41.0

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