Question

At 500°c the reaction between N2 and H2 to form ammonia has Kc= 6.0×10^-2. what is the numerical value of Kp for this reaction? I'm​

Answer 1

Answer:

please mark as

Explanation:

The reaction between N2andH2 to from ammonia has Kc6×10-2 at the temperaturee 500∘C. The numerical value of Kp for this reaction is. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. Kp for the reaction N2+3H2⇔2NH3 is 1.6×10-4atm-2 at 400∘C.

Answer 2

N2+3H2⇌2NH3

Kc=6×10^−2

R=0.0721

T=500°C= 500+273=773K

n=2-3-1=-2

Kp=Kc(RT)-²

Kp=6×10-²(0.0821×773)-²

Kp=6×10-²/(0.0821×773)²

Kp=6×10-²/(63.46)²

Kp=6×10-²/(4027)

Kp=1.5×10-5 (Ans)