For a different reaction, Kc=9.44 x 10^6, Kf= 777s-1, and Kr=8.24 x 10^-5 s-1. A

Question

For a different reaction, Kc=9.44 x 10^6, Kf= 777s-1, and Kr=8.24 x 10^-5 s-1. Adding a catalyst increases the forward rate constant of 6.85 x 10^4 s -1. What is the new value of the reverse reaction constant, Kr, after adding the catalyst? Please provide a step by step process. I thought I had all my math right but guess not. Please and Thank you's in advance ! 3 Mastering chemistry Course Masteringchemistry. Po x O e session masteringchemistry. CHM 1202-01 and CHM 1202-02 Spnng 2017 for McMurry FayR signed in as Ayanna Walker Hrp Post Lecture Homework Chapter 14 t Chemical Equilibrium and Chemical Kinetics previous l 17 of 19 Part B For a dumerent reaction. Ke -944x10e, kn 778s 1.and kr -824x10" s Adding a catalyst increases the forward rate constant to 6.85x10 s What is the new value of me reverse reaction constant, kr, aner adding catalyst? Express your answer with the appropriate units. Include explicit mu within units, for example to enter M s include (multiplication doty between each x, x 10m 8.24.10 s-1 Incorrect; Try Again; 4 attempts remaining predictable change in me other rate constant O Ask me anything

Explanation / Answer

We know for an a reversible reaction

Kc = Kf/Kr

Where Kc is equilibrium constant

Kf is forward rate constant

Kr is reverse rate constant

For a reversible reaction, equilibrium rate constant remains same throughout the reaction but there could be change in forward rate constant and reverse rate constant.

From Question, new Kf = 6.85 x 10^4 s -1

So

Kr = Kf/Kc = (6.85 x 104 s -1) /( 9.44 x 106)

Kr = 0.726 * 10-2 s-1 = 7.26 * 10-3 s-1

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