Please answer question #12. Thank you! 10. The data below was collected for the reaction H, (g) + Br, (g)2 HBr (g), when [H,J=2.0М.[Br,] (M)Rate (M/s)1.3E-032.5E-049.6E-055.3E-053.4E-052.3E-052.5E-051.2E-057.0E-065.3E-064.2E-063.3E-06(Hint: make a log-log plot)Depending on the context,>" may or may not imply that only the forward reaction is signifi-сant.11. The rate law for the reaction SO,Cl, (g)→ SO, (g) + Cl,(g) is:Rate = 2.20 x 105 s-' [SO,CI,]If the reaction starts with 0.222 moles in a 750 mL flask at 320°C, how many moles of SO,Cl, re-main after 1 hour? How many moles of SO, will have been produced? What is the half-life of thisreaction212. Derive the rate law for each mechanism shown below.а.2 I(fast)I+ H,H,I(fast)H,I + IH, + I,2 HI(slow)2 HI(overall)b.NO + Br,NOB.,(slow)NOBr, + NO2 NO + Br,2 NOB1(fast)2 NOB1(overall)Homework Sets36

Answered: question #12

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Please answer question #12. Thank you!

10. The data below was collected for the reaction H, (g) + Br, (g)
2 HBr (g), when [H,J=2.0
М.
[Br,] (M)
Rate (M/s)
1.3E-03
2.5E-04
9.6E-05
5.3E-05
3.4E-05
2.3E-05
2.5E-05
1.2E-05
7.0E-06
5.3E-06
4.2E-06
3.3E-06
(Hint: make a log-log plot)
Depending on the context,
>" may or may not imply that only the forward reaction is signifi-
сant.
11. The rate law for the reaction SO,Cl, (g)
→ SO, (g) + Cl,(g) is:
Rate = 2.20 x 105 s-' [SO,CI,]
If the reaction starts with 0.222 moles in a 750 mL flask at 320°C, how many moles of SO,Cl, re-
main after 1 hour? How many moles of SO, will have been produced? What is the half-life of this
reaction2
12. Derive the rate law for each mechanism shown below.
а.
2 I
(fast)
I+ H,
H,I
(fast)
H,I + I
H, + I,
2 HI
(slow)
2 HI
(overall)
b.
NO + Br,
NOB.,
(slow)
NOBr, + NO
2 NO + Br,
2 NOB1
(fast)
2 NOB1
(overall)
Homework Sets
36

Expert Solution

Step 1

Hi, as you have asked for question (12) only. Hence solving both subparts of 12.

Step 2

(a) Rate law for any reaction is determined by slowest step.

$\begin{array}{rcl} Rate & = & K_{3} (H_{2} I) (I) . . . (1) \\ From the equation - I + H_{2} & \Leftrightarrow & H_{2} I \\ K_{2} & = & \frac{(H_{2} I)}{(H_{2}) (I)} . . . . (2) \\ From equation - I_{2} & \Leftrightarrow & 2 I \\ K_{1} & = & \frac{{(I)}^{2}}{I_{2}} . . . (3) \\ Substituting the values of H_{2} I and {(I)}^{2} from equation (2) and (3) in equation (1) \\ Rate & = & K_{3} (I) K_{2} (H_{2}) (I) \\ = & K_{3} K_{2} (H_{2}) K_{1} (I_{2}) \\ = & K_{3} K_{2} K_{1} (H_{2}) (I_{2}) \\ rate & = & K^{'} (H_{2}) (I_{2}), where K' = K_{3} K_{2} K_{1} \end{array}$

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