Question

Nitrogen dioxide dimerizes according to the following reaction:

2NO2(g)⇌N2O4(g) Kp=6.7at298K A 2.35-L container contains 0.059 mol of NO2 and 0.083 mol of N2O4at298K.

Kc = 164

Calculate Q for the reaction.

Answer 1

According to the Question ,

2NO2 ⇌ N2O4

Kp = 6.7

Temperature  = 298 K

Kp = Kc* (RT)^{del n}

del n = Number of mole Gaseous product - Number of mole Gaseous reactant

=> del n = 1 - 2 = -1

So Now we Calculate = Kc

=> Kc = Kp / ( RT )^-1

=> Kc = Kp * RT

=> Kc = 6.7 * 0.0821 * 298

=> Kc = 163.92

=> Kc = 164 ( ANS -1) You Answer For Kc is correct

Now we Calculate Q

Moles of N2O4 = 0.083 moles

Moles of N2O = 0.059 moles

Volume = 2.35 L

Concentration of N2O4 = Moles / Volume in L = 0.083 / 2.35 = 0.035 M

Concentration of N2O = Moles / Volume in L = 0.059 / 2.35 = 0.025 M

2NO2 ⇌ N2O4

So Now Q = [N2O4] / [N2O]²

=> Q = 0.035 / 0.025²

=> Q = 56

Thank you

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