Question

The average rate of the following reaction at a certain temperature is 3.03×10-2 M/min. Calculate the value for the rate of change of [NaCl], in M/min, at this temperature. Report your answer to three significant figures in scientific notation.

FeCl2(aq) + 2NaOH(aq) → Fe(OH)2(s) + 2NaCl(aq)

If the intial concentration of FeCl2 is 0.891 M and the initial concentration of NaOH is 0.116 M, determine the concentration of NaOH (in M) after 104 seconds. Assume the volume of the solution does not change and report your answer to three significant figures in scientific notation.

Answer 1

The average rate of the given reaaction

FeCl_2(aq) + 2NaOH_(aq) ------> Fe(OH)_2(s) + 2NaCl_(aq)

is expressed as

- d[FeCl2]/dt = -1/2d[NaOH]/dt = d[Fe(OH)2]/dt = 1/2d[NaCl] /dt

The average rate = 3.03x 10^-2M/min given

From the above expression 1/2 d[NaCl]/dt = 3.03x 10^-2M/min

thus the rate of change (formation) of NaCl = 6.06x 10^-2M/min.

  FeCl_2(aq) + 2NaOH_(aq) ------> Fe(OH)_2(s) + 2NaCl_(aq)

0.891 0.116 0 0 initial concentrations(t=0)

? after 104s (t=104/60min)

Rate of change in concentration of NaOH = 2 x average rate (see above discussion)

=- 6.06x 10^-2 M/min

we know average rate = [final concentration - initial concentration] / time taken for change

= [x M - 0.116M ] / (104/60min)

Equating both we get the final concentration = 1.096x 10^-2 M