Part A For a certain reaction, Kc = 6.03×109 and kf= 5.07×102 M2s1 . Calculate t
ID: 491003 • Letter: P
Question
Part A For a certain reaction, Kc = 6.03×109 and kf= 5.07×102 M2s1 . Calculate the value of the reverse rate constant, kr, given that the reverse reaction is of the same molecularity as the forward reaction. Express your answer with the appropriate units. Include explicit multiplication within units, for example to enter M2s1 include (multiplication dot) between each measurement. kr = SubmitHintsMy AnswersGive UpReview Part Part B For a different reaction, Kc = 7.04×105, kf=1.38×105s1, and kr= 0.196 s1 . Adding a catalyst increases the forward rate constant to 3.93×107 s1 . What is the new value of the reverse reaction constant, kr, after adding catalyst? Express your answer with the appropriate units. Include explicit multiplication within units, for example to enter M2s1 include (multiplication dot) between each measurement. kr = SubmitHintsMy AnswersGive UpReview Part Part C Yet another reaction has an equilibrium constant Kc=4.32×105 at 25 C. It is an exothermic reaction, giving off quite a bit of heat while the reaction proceeds. If the temperature is raised to 200 C , what will happen to the equilibrium constant? The equilibrium constant will Yet another reaction has an equilibrium constant at 25 . It is an exothermic reaction, giving off quite a bit of heat while the reaction proceeds. If the temperature is raised to 200 , what will happen to the equilibrium constant? increase. decrease. not change.
Explanation / Answer
nA <-> nB
Rf = Kf*[A]^n
Rr = K*[B]^m
in eq
Rf = R
Kf*[A]^n = k[B]^m
Kf/kr = [B]^m/[A]^n = Kc
Kc = Kf/Kr
so
Kr = Kf/Kc = (5.307*10^-2)/(6.03*10^-9) = 8800995.024
B)
increase in Kf --> 3.93*10^7 /( 1.38*10^5) = 284.7826
so
new K --> 284.7826 * 0.196 = 55.817
C)
if exotheric --> it releases heat
so
inreasing T will not vaour this, actually will decrese the value