What is the value of Kp for the reaction:
NO2(g)+CO(g)->NO(g)+CO2(g)
Given the following information:
2NO2(g)->NO3(g)+NO(g) Kp=2.72x10^(-4)
2NO2(g)+2CO2(g)->2NO3(g)+2CO(g) Kp=5.73
Use eqn 1 as is. Call that Kp = Kp1
Use eqn 2, call it Kp2
Take 1/2 eqn 2 and K becomes sqrt Kp2. Then reverse the 1/2 eqn 2 which now becomes 1/sqrt Kp2.
Tben add in eqn 1. Check to make sure you have the equation you want. Kp for the reaction you want we will call Kp3.
Then Kp3 for the rxn you want will be Kp1*(1/sqrt Kp2) = Kp3
To determine the value of Kp for the given reaction, we need to use the relationship between equilibrium constants for reactions that are the reverse of each other.
Given:
2NO2(g) -> NO3(g) + NO(g) Kp = 2.72x10^(-4)
2NO2(g) + 2CO2(g) -> 2NO3(g) + 2CO(g) Kp = 5.73
Let's set up the reaction in the reverse direction to find the desired Kp value.
Reverse reaction 1:
NO3(g) + NO(g) -> 2NO2(g)
Reverse reaction 2:
2NO3(g) + 2CO(g) -> 2NO2(g) + 2CO2(g)
Now, let's write the equation for the given reaction:
NO2(g) + CO(g) -> NO(g) + CO2(g)
Since we obtained the desired reaction by reversing both reactions, we need to multiply the reverse Kp values to get the Kp value for the given reaction.
Multiply reverse reaction 1 by 2 (to balance the stoichiometry):
2(NO3(g) + NO(g)) -> 4NO2(g)
Reverse reaction 2 remains the same:
2NO3(g) + 2CO(g) -> 2NO2(g) + 2CO2(g)
Now, let's combine the two equations we obtained:
2(NO3(g) + NO(g)) + 2NO3(g) + 2CO(g) -> 4NO2(g) + 2NO2(g) + 2CO2(g)
Simplifying the equation:
2NO3(g) + 2NO(g) + 2NO3(g) + 2CO(g) -> 4NO2(g) + 2CO2(g)
Now, we can cancel out the common terms on both sides of the equation:
2NO(g) + 2CO(g) -> 2NO2(g) + 2CO2(g)
The equation we obtained is the same as the given reaction, so the Kp value for the given reaction is the product of the reverse Kp values:
Kp = Kp(reverse reaction 1) * Kp(reverse reaction 2)
= (2.72x10^(-4)) * (5.73)
= 1.56x10^(-3)
Therefore, the value of Kp for the reaction NO2(g) + CO(g) -> NO(g) + CO2(g) is 1.56x10^(-3).