A flask is charged with 1.220 atm of N2O4(g) and 0.85 atm N02(g) at 25 degree C.

Question

A flask is charged with 1.220 atm of N2O4(g) and 0.85 atm N02(g) at 25 degree C. The equilibrium reaction is given in the equation below. N2O4(g) 2NO2(g) After equilibrium is reached, the partial pressure of NO2 is 0.512 atm. (a) What is the equilibrium partial pressure of N2O4? [ ] x atm (b) Calculate the value of Kp for the reaction. (c) Is there sufficient information to calculate Kc for the reaction? Yes, because the partial pressures of all the reactants and products are specified. Yes, because the temperature is specified. No, because the value of K can be determined experimentally only. If Kc can be calculated, find the value of Kc. Otherwise, enter 0.

Explanation / Answer

given reaction

N204 ---> 2N02

uisng ICE table

inital pressures of N204 , N02 are 1.22 , 0.85

change in pressures of N204 , N02 are -x , +2x

equilibrium pressures of N204, N02 are 1.22-x , 0.85 + 2x

given

partial pressure of N02 at equilirbrium = 0.512

so

0.85 + 2x = 0.512

x = -0.169

so

partial pressure of N204 = 1.22 - x

= 1.22 + 0.169

= 1.389

so

partial pressure of N204 at equilirbium = 1.389 atm

now

2)

N204 ---> 2N02

Kp = (pN02)^2 / (pN204)

using the calculated values

we get

Kp = ( 0.512)^2 / ( 1.389)

Kp= 0.189

so

the value of Kp is 0.189

3)

now

we know that

Kp = Kc(RT)^dn

here

dn = moles of gaseous products - moles of gaseous reactants

N204 ---> 2N02

so

dn = 2 - 1

dn =1

now

KP = Kc (RT)

Kc = Kp/RT

given

T = 25 C = 298 kelvin

R = 0.0821

so

Kc = 0.189 / 0.0821 x 298

Kc = 7.714

so

the value of KC is 7.714

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