# Im stuck on this rate equations question, someone help?Watch

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#1
Rate =k[X][OH-]
Initial concentration of X = 0.024moldm-3
Initial concentration of OH- = 0.035moldm-3
Initial rate of reaction was= 8.5 x 10^-5moldm-3s-1
Rate constant is 0.101 mol-1dm3s-1

QUESTION?
In a second experiment at the same temperature, water was added to the original reaction mixture so that the total volume was doubled. Calculate the initial rate of reaction in this second experiment.
0
2 years ago
#2
The fact that it's at the same temperature tells you k is the same. You know that the orders of the reaction are the same too as it's the same equation. All we have to do now is work out the new concentrations: Let's say originally there was 10cm3 of reaction solution. This then changed to 20cm3.
Moles of X:2.4x10-4
Moles of OH-:3.5x10-4
New concentration:X=0.012 OH-=0.0175 (These are just half of the original concentrations since the reaction mixture volume doubled)
Plugging it into the rate equation:
0.101*0.012*0.0175=2.12x10-5 New initial rate of reaction
Hope this helps
0
2 years ago
#3
Quick way of doing it
2 species in rate equation
Both first order
Both conc will halve when volume doubles
0.5*0.5=0.25
New rate will be a quarter of original value as k is a constant
0
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