Question

A flask that initially contained pure NO2 was heated to 1000 K, a temperature at which the value of Kp for the decomposition of NO2 is 123.7. 2NO2(g) <-----> 2NO(g) + O2(g) Given that the partial pressure of O2 at equilibrium equals 0.1865 atm, calculate the partial pressures of NO2 and NO at equilibrium. PNO = PNO2 =

Name: a flask that initially contained pure no2 was heated to 1000 k a temperature at which the value of kp for the decomposition of no2 is 1237 2no2g 2nog o2g given that the partial pressure of o 35953
Uploaded: 2022-02-28T13:57:04-08:00
Duration: 2 min 51 s
Channel: Krishna G
Description: a flask that initially contained pure no2 was heated to 1000 k a temperature at which the value of kp for the decomposition of no2 is 1237 2no2g 2nog o2g given that the partial pressure of o 35953

          A flask that initially contained pure NO2 was heated to 1000 K,
a temperature at which the value of Kp for the decomposition of NO2
is 123.7.

2NO2(g) <-----> 2NO(g) + O2(g)

Given that the partial pressure of O2 at equilibrium equals
0.1865 atm, calculate the partial pressures of NO2 and NO at
equilibrium.

PNO =
PNO2 =

Added by Alyssa S.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Krishna G

02/28/2022

Step 1

Step 1: Write the expression for the equilibrium constant in terms of partial pressures: \[ K_p = \frac{(P_{NO}^2)(P_{O_2})}{(P_{NO_2})^2} \] Show more…

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A flask that initially contained pure NO2 was heated to 1000 K, a temperature at which the value of Kp for the decomposition of NO2 is 123.7. 2NO2(g) <-----> 2NO(g) + O2(g) Given that the partial pressure of O2 at equilibrium equals 0.1865 atm, calculate the partial pressures of NO2 and NO at equilibrium. PNO = PNO2 =