For the following reaction, A + B + C -> D + E, the rate expression is first ord

Question

For the following reaction, A + B + C -> D + E, the rate expression is first order with respect to reactant A, zero order with respect to reactant B and second order with respect to reactant C. What is the numerical value of the rate constant, k, if the rate = 2.24e-4 mole/lit-min when the [A] = 0.0050M, [B] = 0.250M, and [C] = 0.0112M? B)what are the units of the rate constant K? For the following reaction, A + B + C -> D + E, the rate expression is first order with respect to reactant A, zero order with respect to reactant B and second order with respect to reactant C. What is the numerical value of the rate constant, k, if the rate = 2.24e-4 mole/lit-min when the [A] = 0.0050M, [B] = 0.250M, and [C] = 0.0112M? B)what are the units of the rate constant K? B)what are the units of the rate constant K?

Explanation / Answer

acording to question rate law =

R = K[A]1[B]0[C]2

value of K= (2.24 x 10-4)/(0.0050)(0.250)0(0.0112)2

K= 357.143L2 mol-2 min-1.

unit = (mol/lit-min)/(molL-1)3 = mol-2L2min-1

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