Consider the following gas phase reaction: 3A(g)+ 2B(g) <==> 1C(g)+4D(g). At 181.1C, Kp=270.91. Calculate Kc under these conditions.
My answer:
change of n = (1+4)-(3+2)=0
kp=kc(RT)^(change of of n)
207.91=kc(.0821)(454.1)^0
Therefore,
kc=270.91
Is this correct?
yes, Kp=Kc because delta n = 0
Yes, your answer is correct! In the equation you provided, the change in the number of moles, Δn, of the gaseous reactants and products is zero (1+4)-(3+2) = 0. Since Δn = 0, the equilibrium constant in terms of pressure, Kp, is equal to the equilibrium constant in terms of concentration, Kc.
To solve for Kc, you need to use the following equation:
Kp = Kc * (RT)^(Δn)
R = 0.0821 L*atm/(mol*K) is the ideal gas constant, and T = 181.1°C = 454.1 K is the temperature. Since Δn is zero, the equation simplifies to:
Kp = Kc * (RT)^0
Now, (RT)^0 equals 1. So, the equation becomes:
Kp = Kc * 1
Since Kp = 270.91, we can set Kc equal to Kp:
Kc = Kp
Therefore, Kc = 270.91.