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A flask is charged with 1.800atm of N2O4(g) and 1.00 atm NO2(g) at 25 ?C, and...

A flask is charged with 1.800atm 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 0.512atm . Calculate Kc for the reaction.

Homework Answers

Answer #1


                            N2O4(g)<------>2NO2
initial                        1.8                   1
at equilibrium            1.8-x                1+2x

given partial pressure of NO2= 0.512 atm
so,1+2x=0.512
x= -0.244

p(N2O4) = 1.8 - x = 1.8 + 0.244 = 2.044 atm

Kp = 0.512 ^2 / 2.044
    =0.1283

Kc = Kp /(R*T)^delta n
    = 0.1283 / (0.0821 * 298)^1
   =5.242 *10^-3

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