Consider the reaction 2 N2O5 <=> 4NO2 + O2 Some N2o5 is placed in a container, e

Question

Consider the reaction 2 N2O5 <=> 4NO2 + O2

Some N2o5 is placed in a container, exerting a pressure of 1.00 atm before any reaction. When equlibirum is reached, the total pressure in the container is 2.00 atm. Calculate the Kp for this reaction.

*I set up the ice table and I got ((4x)^4)x(x)) / ((1-2x)^2)) and i do not know where to go from there

Part 2

A container is initially filled with .250 atm of CO and .0450 atm of O2. Calculate the pressure of O2 after equilibrium is attained.

2 CO + O2 => 2CO2 Kp=6.0x10^8

Do not know how to approach

Explanation / Answer

a) do this qusetion like this

since pressure was 1 before everything so it must be of N2O5

intially

2 N2O5 <=> 4NO2 + O2

1                         0         0

after equilibrium

2 N2O5 <=> 4NO2 + O2

1-2p               4p            p

total pressure =1-2p+4p+p =1+3p =2

so p=1/3

so

after equilbrium what is the condition

2 N2O5 <=> 4NO2 + O2

1/3                   4/3          1/3

so Kp =(4/3)^4*(1/3)/(1/3)^2 =9.48 atm^3

b)similarly

intitallly

2 CO + O2 => 2CO2

0.25     0.0450      0

after

2 CO + O2 => 2CO2

0.25-2p     0.045 -p     2p

Kp =(2p)^2/[(0.25-2p)^2(0.045 -p )]

solve to get p

and answer is 0.045 -p

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