How would I answer part b.ii in this rate equations question?

(b) The reaction between two different compounds, C and D, is studied at a given
temperature.
The rate equation for the reaction is found to be
rate = k[C][D]
2
(i) When the initial concentration of C is 4.55 × 10–2 mol dm–3 and the initial
concentration of D is 1.70 × 10–2 mol dm–3, the initial rate of reaction is 6.64 ×
10–5 mol dm–3 s–1
.
Calculate the value of the rate constant at this temperature and deduce its
units.
Calculation ............................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
Units of rate constant ............................................................................
...............................................................................................................
(3)
(ii) The experiment in part (i) is repeated at the same temperature but after the
addition of extra solvent so that the total volume of the mixture is doubled.
Deduce the new initial rate of reaction.

I understand how to calculate the first part but I'm unsure of how to do part ii

Reply 1

Nitrotoluene

Original post

by Slugzie(:

(b) The reaction between two different compounds, C and D, is studied at a given
temperature.
The rate equation for the reaction is found to be
rate = k[C][D]
2
(i) When the initial concentration of C is 4.55 × 10–2 mol dm–3 and the initial
concentration of D is 1.70 × 10–2 mol dm–3, the initial rate of reaction is 6.64 ×
10–5 mol dm–3 s–1
.
Calculate the value of the rate constant at this temperature and deduce its
units.
Calculation ............................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
Units of rate constant ............................................................................
...............................................................................................................
(3)
(ii) The experiment in part (i) is repeated at the same temperature but after the
addition of extra solvent so that the total volume of the mixture is doubled.
Deduce the new initial rate of reaction.
I understand how to calculate the first part but I'm unsure of how to do part ii

Hello Slugzie(:!

Part ii: When you double the total volume of the mixture, the concentrations of C and D decrease to half. That happens because the total amount of reactants is constant while the volume of the mixture increases.
The new concentrations of C and D are:
[C] = 4.55 × 10^-2 mol/dm^3 / 2 = 2.275 × 10^-2 mol/dm^3
[D] = 1.70 × 10^-2 mol/dm^3 / 2 = 8.50 × 10^-3 mol/dm^3
Using the rate equation again, we can calculate the new initial rate of the reaction:
Rate = k × ([C] × [D]) = (8.59 × 10^-3 mol^-1 dm^3 s^-1) × (2.275 × 10^-2 mol/dm^3) × (8.50 × 10^-3 mol/dm^3) = 1.73 × 10^-5 mol/dm^3 x s
Step by step:
(8.59 × 10^-3 mol^-1 dm^3 s^-1) × (2.275 × 10^-2 mol/dm^3) = 19.53 × 10^-5 mol^-1 dm^3 s^-1
Then, multiplying by the third factor:
19.53 × 10^-5 mol^-1 dm^3 s^-1 × (8.50 × 10^-3 mol/dm^3) = 165.91 × 10^-8 mol/dm^3 s^-1
So, the new initial rate of reaction is 165.91 × 10^-8 mol/dm^3 s^-1 or 1.6591 × 10^-6 mol/dm^3 s^-1

Bye,
Sandro Italy Flag.png