For a certain reaction, Kc = 265 and kf= 7.19 M2s1 . Calculate the value of the
ID: 539974 • Letter: F
Question
For a certain reaction, Kc = 265 and kf= 7.19 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.
Part B For a different reaction, Kc = 1.14×106, kf=51.7s1, and kr= 4.52×105 s1 . Adding a catalyst increases the forward rate constant to 1.82×104 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.
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? Hints 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
1.for a raacttion, equilibrium is a point at which the rate of forward reaction = rate of backward reaction
Equilibrium constant, Kc= Forward reaction rate constant/ backward reaction rate constant = kf/Kr
Kr= Kf/Kc= 7.19/265= 0.027 /M2.s
2. for the reaction given KC= Kf/Kr= 51.5/(4.52*10-5)= 11.4*105
the presence of catalyst does not influence the equilibrium constant. It increases the rate of forward reaction as well as backward reaction. Hence , KC remains the same
hence even in the presece of catalyst, KC= 11.4*105= 1.82*104/Kr
Kr= 1.82*104/ (11.4*105)= 0.0159
3. Kc= 4.32*105 at 25 deg.c, As per Van;t Hoff equation, ln (K200/K25)= (deltaH/R)*(1/T1-1/T2)
where T1= 298K, T2= 200+273= 473K, K200= equilibrium constant at 200 deg,c (473K), T1= 25 deg.c (298K)
K200= equilibrium cosntant at 200 deg.c, K25= equilibrium constant at 25 deg.c
deltaH= enthalpy change of reaction, ln (K200/K25)= (deltaH/R)*(1/298-1/473)= (deltaH/R)*0.001242
since for exothermic reaction, deltaH is -ve, K200/K25 will be <1, K200 will decrease