1- The following data were obtained for 2A + 2B + C----->D+E: Trial [A] mol/L [B

Question

1- The following data were obtained for 2A + 2B + C----->D+E:

Trial         [A] mol/L       [B] mol/L          [C] mol/L                           Rate mole/L/sec

1                0.151               0.151              0.151                                    0.48

2               0.302               0.151             0.151                                      0.96

3               0.151               0.151             0.076                                       0.12

4.              0.151               0.302             0.151                                     1.92

A)   what is the rate equation and what is K?

B) If you are trying to achieve a rate of 1.5 mol/L/sec, and the initial concentration of A and were both 0.200M, what would the initial concentration of C have to be?

Explanation / Answer

since the rate double by doubling the concentration of A, it is first order wrt A

similarly doing for B and C, we get that it inversely proportional to [C]^2

also,

it is 2nd order with respect to [B]

so rate=k[A][B]^2/[C]^2

so,

0.48=k*0.151

or k=3.18 s^-1

B)1.5=3.18*0.2^2*0.2/x^2'

or x=[C]=0.13 M

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.