CONCEPTUAL EXERCISES Exercises 87 and 88 are designed to simulate arrival at equ
ID: 812393 • Letter: C
Question
CONCEPTUAL EXERCISES Exercises 87 and 88 are designed to simulate arrival at equilibrium through a series of steps.The instructions indicate how to get from one step to the next. Record the number of each type of molecule present at each step. Relate the number of molecules of each type at equilibrium to the prescription for converting molecules at each step. 87. Examine a prototypical reaction ASB. a. Start with 50 molecules of A in 1 L. The reaction prescrip- tion is as follows. At each step, 0.1 of the A molecules present react to yield B molecules, and 0.005 of the B molecules present react to form A molecules. What are the equilibrium concentrations? How many cycles does it take to arrive at equilibrium? b. Run the reaction again starting with 50 B molecules and no A molecules. How do the equilibrium concentrations differ? How many cycles does it take to get to equilib- rium? Compare the result with that from part (a) 88. In this simulation, the reaction generates two products: At each step, let one-tenth of the A molecules be transformed into B and C. Because B and C need to come together to the fraction that is transformed involves the product A, form of the concentration of B times that of C. Set the fraction of molecules that are transformed into A molecules x ICJ, to 0.025. Start with 100 molecules of A and 10 molecules o C. What are the equilibrium concentrations? How many cycles does it take to arrive at equilibrium?Explanation / Answer
a) Molecule A completely decomposes into B in 500 steps, and in 500 steps, 2.5 molecules of B decomposes into A
This 2.5 molecules of A takes another 25 steps to decompose to B, and in 25 steps, 0.125 molecules of B decomposes
This 0.125 molecules of A decemposes into B in 1 step and in 1 step 0.005 molecules of B decomposes.
At the end, 0.03 molecules of A and (47.5+2.375+0.095) = 49.97 molecules of B remains, at equilibrium
No. of cycles or steps = 526. Decomposition of B continues to form A until A reaches 0.1, that is another 0.07/0.005 = 14 steps.
Thereofre, total number of steps = 540 and the equilibrium concentration of A is 0.1 and B is 49.9
b) No. of steps required to reach 0.1 molecules of A is 0.1/0.005 = 20.
and once the number of molecules of A reaches just below 0.1, the reaction virtually stops or reaches equilibrium
Equilibrium concentration of A is 0.1 and B is 49.9, number of steps = 20
88)
B and C - consider them as one product as BC, For calculation purpose the ratio of [B]/[C] does not really matter.
number of steps 100 molecules of A to decompose completely into B &C is 1000, and in 1000 steps 25 molecules of B & C decomposes
25 molecules decomposes into B&C in 250 steps and in 250 steps 6.25 molecules of A is formed
in the next 62 steps, 1.55 molecules of A is formed and 0.05 remains from previous step.
1.6 molecules of A takes 16 steps to form products and the products decomposes to form 0.4 molecules of A
0.4 molecules of A takes 4 steps to form products and the products form 0.1 molecules of A.
Here the reaction reaches equilibrium
Thus the concentration of A at equilibrium is 0.1 and the products B&C together 99.9 molecules
Number of cycles to reach equilibrium = 1000 + 250 + 62 + 16 + 4 = 1332 steps.
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