Consider the reaction A+2BC whose rate at 25 C was measured using three differen

Question

Consider the reaction

A+2BC

whose rate at 25 C was measured using three different sets of initial concentrations as listed in the following table:

What is the rate law for this reaction?

Express the rate law symbolically in terms of k, [A], and [B].

The activation energy of a certain reaction is 38.2 kJ/mol . At 26  C , the rate constant is 0.0120s1. At what temperature in degrees Celsius would this reaction go twice as fast?

Express your answer with the appropriate units.

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Part B

Given that the initial rate constant is 0.0120s1 at an initial temperature of 26  C , what would the rate constant be at a temperature of 200  C for the same reaction described in Part A?

Express your answer with the appropriate units.

Trial [A]
(M) [B]
(M) Rate
(M/s) 1 0.40 0.010 1.9×103 2 0.40 0.020 3.8×103 3 0.80 0.010 7.7×103

Explanation / Answer

Part B:

Given:

K1=0.0120 s-1

Temp. T1=26C =299 K

Ae=31.6 kJ/mol=31600 J/mol

T2=200 ? = 473 K

From arrhenius equation:

ln(K2/K1­)=Ea/R (1/T1 -1/T2)

ln(K2/0.012)=31600/8.314 *(1/299 -1/473)

K2=1.08 s-1

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