A flask is charged with 1.350atm of N2O4(g) and 1.00 atm NO2(g) at 25 ∘C, and the following equilibrium is ach?
A flask is charged with 1.350atm of N2O4(g) and 1.00 atm NO2(g) at 25 ∘C, and the following equilibrium is achieved:
N2O4(g)⇌2NO2
After equilibrium is reached, the partial pressure of NO2 is 0.514atm .
1)What is the equilibrium partial pressure of N2O4?
2)Calculate the value of Kp for the reaction.
3)Calculate Kc for the reaction.
please explain, thanks :)
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The ideal gas law is pV=nRT. Since V and T are the same for the two gases (both initially and at equilibrium), the partial pressures are in the same ratio as the numbers of moles (or molecules).
So suppose we start with 1.350 mol of N2O4 and 1.00 mol of NO2. At equilibrium we have 0.514 mol of NO2, so 1.000-0.514=0.486 mol of NO2 has combined to form 0.486/2=0.243 mol of N2O4. So at equilibrium we must have 1.350+0.243=1.593 mol of N2O4. The partial pressure of N2O4 at equilibrium has increased in proportion to the number of moles present, so it must be 1.593 atm.
Kp = p(NO2)^2/p(N2O4)
= 0.514^2/1.593
= 0.166 atm (rounded to 3 sig. figs, the smallest used in the question).
Kc = [NO2]^2/[N2O4]
at equilibrium, measured in mol/dm3.
Since pV=nRT then n/V=p/RT. Since Kp is in units of pressure, to convert Kp to Kc we need to divide by RT while being careful with units.
0.166 atm = 0.166 x 101,325 Pa = 16,800 N/m2.
R=8.314 J/molK=8.314 Nm/molK.
T=25+273=298K.
RT=8.314 Nm/molK x 298 K
=2478 Nm/mol.
p/RT=(16800 N/m2) / (2478 Nm/mol)
= 6.78 mol/m3
= 6.78 x 10^-3 mol/dm3
since 1 m3 = 1000 dm3.
1.35