A flask is charged with 1.760 atm of N 2 O 4 ( g ) and 1.07 atm NO 2 ( g ) at 25

Question

A flask is charged with 1.760 atm of N2O4(g) and 1.07 atm NO2(g) at 25?C. The equilibrium reaction is given in the equation below.

After equilibrium is reached, the partial pressure of NO2 is 0.512 atm.

(a) What is the equilibrium partial pressure of N2O4?
atm
(b) Calculate the value of Kp for the reaction.
If Kc can be calculated, find the value of Kc. Otherwise, enter 0.

A flask is charged with 1.760 atm of N2O4(g) and 1.07 atm NO2(g) at 25½C. The equilibrium reaction is given in the equation below. N2O4(g) A flask is charged with 1.760 atm of N2O 2NO2(g) After equilibrium is reached, the partial pressure of NO2 is 0.512 atm. (a) What is the equilibrium partial pressure of N2O4? atm (b) Calculate the value of Kp for the reaction.

Explanation / Answer

... ... ... initial ... ... change ... ... final
NO2 . 1.07 atm ... .. -2x ... ... 0.512 atm
N2O4 1.760 atm ... ... +x ... ... 1.760+x atm

(a)
-2x = 0.512 - 1.07 atm
x = 0.279 atm

final N2O4:
1.760 + 0.279 = 2.039 atm

(b)
Kp = [p(NO2)]

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