the reaction
A+B⟶C+Drate=k[A][B]2
has an initial rate of 0.0450 M/s.
What will the initial rate be if [A] is halved and [B] is tripled?
initial rate:
M/s
What will the initial rate be if [A] is tripled and [B] is halved?
initial rate:
how do i go about this?
Sapling Question
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Re: Sapling Question
If [A] is tripled and [B] is halved
Rate = k[A][B]2
rate = k[3][0.5]2
rate = 0.75k which means that the rate will be 0.75 (three quarters) the original rate (since k is constant)
Initial rate under these conditions will be 0.75 x 0.0450 M/s = 0.034 M/s
Rate = k[A][B]2
rate = k[3][0.5]2
rate = 0.75k which means that the rate will be 0.75 (three quarters) the original rate (since k is constant)
Initial rate under these conditions will be 0.75 x 0.0450 M/s = 0.034 M/s
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Re: Sapling Question
Rule of thumb for questions like these would be to make sure you have correctly assigned orders to the reactants (have the right rate law). This can then be used to determine the effect on rate based on changes to concentrations. In your example of [A][B]^2, you can calculate how the changes would affect the rate law by putting the changes into your equation and multiplying it against your original rate that is given to you. You just want to make sure that you're not leaving out the exponents from your orders that you assign.
first order= change will have the same change on rate (ex. tripling concentration will triple the rate)
second order= change will be squared then applied to rate (ex. halving concentration -> (0.5)^2 = 0.25 so the rate will be 1/4 of what is was)
another example for this would be doubling concentration -> 2^2 = 4 so rate will be four times what it was
applying the exponents then looking at the effect gives you some more insight on zero order reactants
zero order= no change (ex doubling concentration -> 2^0 = 1 so rate will not change)
because there is no change, this is why zero order reactants are not included into your law
first order= change will have the same change on rate (ex. tripling concentration will triple the rate)
second order= change will be squared then applied to rate (ex. halving concentration -> (0.5)^2 = 0.25 so the rate will be 1/4 of what is was)
another example for this would be doubling concentration -> 2^2 = 4 so rate will be four times what it was
applying the exponents then looking at the effect gives you some more insight on zero order reactants
zero order= no change (ex doubling concentration -> 2^0 = 1 so rate will not change)
because there is no change, this is why zero order reactants are not included into your law
Re: Sapling Question
hi! so basically this question gives you the rate law which is k [A][B]^2. when [A] is halved, it would become [1/2A] and when B is tripled it would become [3B]^2. this would then become 1/2[A] and 9[B] because 3 is squared. When you write the whole thing together, you would get rate = 9k[A][B]/(2) which would mean 9/2*(k[A][B]^2) = 9/2*initial rate. you would do the same thing for the second part with numbers reversed. i hope this makes sense :)
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