5 5 At 25°C, N:04(g) decomposes as follows: N20(g)2NO(g) and K-4.61 x 10 If 2.6atm N204(g)...

Question

Question

5

Answer 1

Answer #1

The relationship between kp and kc is as follows:

kp = kc (RT)^∆n

∆n is the difference in number of moles in product side and the number of moles in reactant side.

∆n = 2-1 = 1

Kp= kc* (0.08206*298) = 4.61*10^-3*

= 0.1127

Kp = p(NO2)^2/p(N2O4) = (2x)^2/(2.6-x)

0.1127(2.6-x) = 4x^2

0.293 - 0.1127x = 4x^2

x = 0.2569

Equilibrium total pressure = 2.6-x + 2x = 2.6+x

= 2.6+0.2569 = 2.8569 atm

Answer: 2.8569 atm

Answer 2

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