Question
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
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