Part A) For a certain reaction, Kc = 2.48×109 and kf= 6.80×102 M2s1 . Calculate
ID: 501699 • Letter: P
Question
Part A) For a certain reaction, Kc = 2.48×109 and kf= 6.80×102 M2s1 . Calculate the value of the reverse rate constant, kr, given that the reverse reaction is of the same molecularity as the forward reaction.
Part B)For a different reaction, Kc = 3.64×103, kf=703s1, and kr= 0.193 s1 . Adding a catalyst increases the forward rate constant to 2.18×104 s1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?
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? Will the equilibrium increase, decrease, or not change?
Explanation / Answer
Part A) For a certain reaction, Kc = 2.48×109 and kf= 6.80×102 M2s1 . Calculate the value of the reverse rate constant, kr, given that the reverse reaction is of the same molecularity as the forward reaction.
Kc = Kf / Kr
2.48×10^9 = 6.80×10^2 / Kr
Kr = 2.74 x 10^7 M2s1
reverse rate constant, kr = 2.74 x 10^7 M2s1
Part B)For a different reaction, Kc = 3.64×103, kf=703s1, and kr= 0.193 s1 . Adding a catalyst increases the forward rate constant to 2.18×104 s1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?
Kc = Kf / Kr
3.64×10^3 = 2.18×10^4 s1 / Kr
reversse rate constant Kr = 5.99 s-1
Part C)
equilibrium constant decreases.
because this is exothermic reaction . so if temperature is raised then the reaction shifts to back ward direction. then Kc decreases