2. Chemical Equilibrium Chemical Equilibrium and Chemical Kinetics 5 of 21 Part

Question

2. Chemical Equilibrium Chemical Equilibrium and Chemical Kinetics 5 of 21 Part A For a certain reaction, Ke 6.93 and k90M Express your answer numerically in inverse seconds. sCalculate the value of the reverse rate constant, k Learning Goal: To understand the relationship behveen the equilibrium constant and rate constants For a general chemical equation the equillbrium constant can be expressed as a ratio of the View Available Hint(s) A +B C+D concentrations CAB If this is an elementary chemical reaction, then there is a single forward rate and a single reverse rate for this reaction, which can be written as follows Submit forward raterAB reverse rate kCD Part B where kr and k are the forward and reverse rate constants respectively. When equilibrium is reached, the forward and reverse rates are equal For a different reaction, Ke1.03 x 10, 9.23 x 10s, and k 9.00x10-s1Adding a catalyst increases the forward rate constant to 4.13 108 s What is the new value of the reverse reaction constant, k, after adding catalyst? Express your answer numerically in inverse seconds. View Available Hint(s) Thus, the rate constants are related to the equilibrium constant in the following manner: c [A][B]

Explanation / Answer

Part a

Kc = 6.93

kf = 9.00*10^-2 M2/s

Reverse rate constant kr = kf / Kc

kr = 9.00*10^-2 / 6.93

kr = 0.01298 M2/s

Or

kr = 1.298*10^-2 M2/s

Part b

Kc = 1.03*10^8

kf = 9.23*10^5 s-1

kr = 9*10^-3 s-1

After adding catalyst equilibrium constant remains the same

kf'' = 4.13*10^8 s-1

New value of reverse rate constant

kr'' = kf"/Kc

= 4.13*10^8/1.03*10^8

= 4.01 s-1

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