given the equilibrium constant for the following reaction at 500 K,
2 NO (g) + O2 (g) <==> 2 NO2 (g) Kc = 6.2x10^5
calculate the equilibrium constant for the reaction expressed as partial pressures, Kp.
the answer is 1.51x10^4. How do i get this?
Kp = Kc(RT)delta n
delta n = nproducts-nreactants.
hhk
To calculate the equilibrium constant (Kp) expressed as partial pressures, you need to use the ideal gas law and consider the expression for Kp in terms of partial pressures.
The expression for Kp is as follows:
Kp = (P(NO2)^2) / (P(NO)^2 * P(O2))
In this equation, P(NO2), P(NO), and P(O2) represent the partial pressures of the corresponding gases at equilibrium.
Given the equilibrium constant (Kc) for the reaction is 6.2x10^5, we can use the relationship between Kc and Kp to find Kp.
The relationship between Kc and Kp is as follows:
Kp = Kc * (RT)^(∆n)
Where R is the ideal gas constant (0.0821 L · atm/(mol · K)), T is the temperature in Kelvin (500 K), and ∆n represents the change in moles of gaseous products minus the change in moles of gaseous reactants (2-3=-1).
Now, let's calculate Kp using the given values:
Kp = Kc * (RT)^(∆n)
= (6.2x10^5) * (0.0821 L · atm/(mol · K))^(-1) * (500 K)^(-1)
= 6.2x10^5 * (1/0.0821) * (1/500)
= 6.2x10^5 * 12.16 * 0.002
= 1.511x10^4
Therefore, the equilibrium constant for the reaction expressed as partial pressures (Kp) is 1.511x10^4.