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Nola504

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Determine the numerical value of the rate constant for the reaction A+B⟶C from the experimental data given below.

Experiment [A] Rate
1 0.43 M 0.21 M 1.52 M/s
2 0.43 M 0.44 M 6.06 M/s
3 0.87 M 0.21 M 3.05 M/s

A:k=(0.45)(0.21)1.52

B: k=1.52(0.43)(0.21)2


C: k=(0.43)(0.21)21.52

D: k=(0.21)21.52

E: k=(0.43)2(0.21)21.52

Explanation
The data in the table shows is the effect of doubling each reactant on the overall rate of reaction. When [A] is doubled, the rate of reaction also doubles. Thus, the reaction is first order with respect to A. When is doubled the rate is quadrupled. Thus the rate is 2nd order in B. The rate constant expression is rate=k[A]2 . If we solve this for k and plug in the values for Rate, [A], and from the first experiment, we get the correct answer. (Note that the values for any of the experiments can be used; the value of k will not differ other than by experimental error)


I think this is an error. A should not even be in the equation. Please help! i am so confused.
 
why do you think that A should not be in the rate law?

rate=k[A]^x^y

in the data provided, when B is kept constant, and the [A] is doubled (.43 x 2 roughly equals .87). If you look at the rates, when A doubles and B is held constant, the rate doubles (1.52 --> 3.05).
Thus the order for A is 1
making the rate law

rate=k[A]^1^2 rearrange and solve and you get your answer
 
why do you think that A should not be in the rate law?

rate=k[A]^x^y

in the data provided, when B is kept constant, and the [A] is doubled (.43 x 2 roughly equals .87). If you look at the rates, when A doubles and B is held constant, the rate doubles (1.52 --> 3.05).
Thus the order for A is 1
making the rate law

rate=k[A]^1^2 rearrange and solve and you get your answer



I don't think any of the answer is in the choices. If you rearrange to solve for K, there should be a division sign.
 
O ya you r right. I totally overlooked that. Thanks. BRZ24, there is a division sign suppose to be after 1.52 in choice B. I guess it didn't copy over. THank you guys for ur help. 🙂
 
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