Answer to Question #69175 in General Chemistry for Ram

Question

When 0.500 mol of N2O4 is placed in a 4.00 L reaction vessel and heated up to 400 K, 79.0% of the N2O4 decomposes to NO2. Calculate Kc and Kp for this reaction

Answer 1

Ideal gas law: pV = nRT; => p = nRT/V

P is the pressure of the gas
V is the volume of the gas (1 liter = 0.001 m^3)
n is the amount of substance of gas (in mol),
R is the ideal, or universal, gas constant, equal to the product of the Boltzmann constant and the Avogadro constant,
T is the absolute temperature of the gas

T = 400 K
V = 4 liter = 0.004 m^3
n(first) = 0.5 mol
w = 79% = 0.79

N2O4 = 2 NO2

Behind the equation of the reaction, from 1 mol N2O4 come out 2 mol NO2.
If reacted 79% N2O4 , than we have: 0.500*((1-0.79)+ 2*0.79) = 0.895 mol (all gas)
p = nRT/V
p = 0.895*8.31*400 / 0.004 = 743745 (Pa)

Partial pressure:
N2O4
1.79(all gas) - 743745 Pa
0.21(N2O4 not reacted) - x Pa
x = 0.21*743745/1.79 = 87255 Pa

NO2
1.79(all gas) - 743745 Pa
1.58(NO2) - x Pa
x = 656490 Pa

Kp = (p(NO2))^2 / p(N2O4)
Kp = 4939305 = 4.94*10^6

n/V = c
p = nRT/V = cRT
Kc = (c(NO2))^2 / c(N2O4)
Kp = (cRT(NO2))^2 / cRT(N2O4) = (c(NO2))^2 * RT/ c(N2O4)
Kp = Kc*RT
Kc = Kp/RT
Kc = 4939305 / 8.31*400 = 1486 = 1.49*10 ^3

Kp = 4939305 = 4.94*10^6
Kc = 1486 = 1.49*10 ^3