For the Reaction A(g)+B(g)+C(g)->D(g) the following data were obtained at constant temperature.

Question

Expt:

1– initial rate=6.25x10^-3 M/s, initial [A]= 0.0500 M, initial [B]= 0.0500 M, initial [C]= 0.0100 M

2– initial rate=1.25x10^-2 M/s, initial [A]= 0.1000 M, initial [B]= 0.0500 M, initial [C]= 0.0100 M

3– initial rate=5.00x10^-2 M/s, initial [A]= 0.1000 M, initial [B]= 0.1000 M, initial [C]= 0.0100 M

4– initial rate=6.25x10^-3 M/s, initial [A]= 0.0500 M, initial [B]= 0.0500 M, initial [C]= 0.0200 M

a) what is the order with respect to each reactant?

b) write the law

c) calculate k (using data from experiment 1)

d) using the value of k calculated in part c, calculate the rate when [A]=0.2000 mol/L, [B]= 0.1500 mol/L, [C]= 0.0300 mol/L

Answer 1

Great answer!

One little extra

To determine each component order you can use the equation

Rate2/rate1= (conc2/conc1)ⁿ

^{n would be the order of that particular reactant}

^{For example in expt1 and expt2, B and C concentration are held constant.}

^{Therefore we can see how doubling the concentration of A alone effects the rate.}

^{if you enter the rate changes from expt2 divided by the rate of expt1 and set equal to concA expt2 divided by concA expt1 raised to the n power}

^{If we solve for n, n=1}

Answer 2

(a) To find the order with respect to each reactant, we want to compare experiments where the reactant of interest changes concentration while the others remain constant.

For A: compare 1 and 2. [A] doubles, all others are constant. Rate also doubles (from 6.25x10^-3 to

1.25x10^-2). This tells us the reaction is FIRST ORDER with respect to A.

For B: compare 2 and 3. [B] doubles, all others are constant. The rate increases 4 fold (from 1.25x10^-2 to 5x10^-2). This tells us the reaction is SECOND ORDER with respect to B.

For C: compare 1 and 4. [C] doubles, all others are constant. Rate remains the same at 6.25x10^-3. This tells us the reaction is ZERO ORDER with respect to C.

(b) The rate law is written as follows:

rate = k[A][B]² .. note that the [C] is not included as it is zero order and rate in independent of [C].

(c) Use the data from experiment one, and plug values into the rate law, and solve fo k.

rate = k[A][B]²

6.25x10^-3 M/s = k[0.05 M][0.05 M]²

6.25x10^-3 M/s = 2.5x10^-3 M³ k

k = 2.5 M^-2s^-1

(d) rate = k[A][B]²

rate = 2.5 M^-2s^-1[0.200 M][0.150 M]²

rate = 0.0113 M/s

Be sure to check all the math. Hope this helps.