In this question, the experiment is repeated with double volume- how would this affect the inital rate of reaction? I have the concs of both reactants, value of K and the old initial rate of reaction
k=5.05 and other values listed in question. Thank you (it's the second part of the question I'm stuck on)
In this question, the experiment is repeated with double volume- how would this affect the inital rate of reaction? I have the concs of both reactants, value of K and the old initial rate of reaction
k=5.05 and other values listed in question. Thank you (it's the second part of the question I'm stuck on)
The rate of reaction (initial or not) depends upon the concentration of reactants C and D, hence their concentrations are shown in the rate equation. Repeat the experiment with different starting concentrations and the rate of reaction will be different too - it must be if you look at the rate equation for this reaction.
If you double the volume using just solvent, you will halve the concentrations of C and D. Work those concentrations out and put those values in the rate equation. K will be unchanged as the temperature will be the same in the repeated experiment.
The rate of reaction (initial or not) depends upon the concentration of reactants C and D, hence their concentrations are shown in the rate equation. Repeat the experiment with different starting concentrations and the rate of reaction will be different too - it must be if you look at the rate equation for this reaction.
If you double the volume using just solvent, you will halve the concentrations of C and D. Work those concentrations out and put those values in the rate equation. K will be unchanged as the temperature will be the same in the repeated experiment.
Ahh thank you, I have the answer now- I was doubling the concentrations rather than halving them. Could you explain a bit more as to why doubling the volume halves the concentration.. is it to do with concentration = moles / volume.. so if the volume is 2 times as big it will cause the concentration to half?
Ahh thank you, I have the answer now- I was doubling the concentrations rather than halving them. Could you explain a bit more as to why doubling the volume halves the concentration.. is it to do with concentration = moles / volume.. so if the volume is 2 times as big it will cause the concentration to half?