Calculate the equilibrium constant Kp for the reaction at a temperature of 298 K.?

Calculate the equilibrium constant Kp for the reaction at a temperature of 298 K.

first find delta G for the reaction:
product-reaction:
{ 2(-228)}product – {-300.4+2(-33.01)}=90.78kj/mol >>>> x1000 =90780 joules
use the formula : G= -RTlnK
k=8.18 x 10^15

2NH3(g) ? N2(g) + 3H2(g) preliminary moles of NH3 = x moles of N2 = (a million/2) * 0.9938x moles of H2 = (3/2)* 0.9938x moles unreacted of NH3 = (a million – 0.9938)x entire moles in equilibrium state = (a million/2) * 0.9938x + (3/2)* 0.9938x + (a million – 0.9938)x = a million.9938 x Kp = (p N2)(p H2)^3/(p NH3)^2 Kp = ((a million/2) * 0.9938x * a million/(a million.9938 x) * 3.28)((3/2)* 0.9938x * a million/(a million.9938 x) * 3.28)^3/((a million – 0.9938)x * a million/(a million.9938 x) * 3.28)^2 Kp = 115888.sixty 4

2NH3(g) ? N2(g) + 3H2(g) preliminary moles of NH3 = x moles of N2 = (a million/2) * 0.9938x moles of H2 = (3/2)* 0.9938x moles unreacted of NH3 = (a million – 0.9938)x total moles in equilibrium state = (a million/2) * 0.9938x + (3/2)* 0.9938x + (a million – 0.9938)x = a million.9938 x Kp = (p N2)(p H2)^3/(p NH3)^2 Kp = ((a million/2) * 0.9938x * a million/(a million.9938 x) * 3.28)((3/2)* 0.9938x * a million/(a million.9938 x) * 3.28)^3/((a million – 0.9938)x * a million/(a million.9938 x) * 3.28)^2 Kp = 115888.sixty 4

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