2. Let\'s apply the modeling technique developed in Question 1 to a reversible r

Question

2. Let's apply the modeling technique developed in Question 1 to a reversible reaction, a) In each second (exchange step), allow 10% of A to react to form B, then allow 10% of B to react to form A. Record the results in the first two columns of the table below. b) In each second (exchange step), allow 10% of A to react to form B, then allow 5% of B to react to form A. Record the results in the last two columns of the table below. 10%/ 10% 10%/5% 0. 100 100 0 24 31 69 6 63 3 3 40 42 L43 10 12 13 14 15 56 s6 4S

Explanation / Answer

As clear from the table

1. let we have 100 units of A out of these 100 units 10 will react to give 10 of B, then out of 10 of B (10%) 1 will react to give back A so after one second we have 91 of A and 9 of B.

2. In next step again 10% of 91 (9.1) will react to form 9.1 of B . out of these 18.1 ( 9 + 9.1) 10% will react back (1.81) to give back A

so we have 83.71 ( approx 84) of A and 16.29 ( approx 16) of B

and so on.....

In Method B

1. Again Let we have 100 units of A and 0 units of B initially

Out of 100 units of A 10 will react to give 10 of B. Out of 10 of B ,5% (0 .5) will react back to give A

so we have 90.5 of A and 9.5 of B

similarly again 10% of A will react and so on ...

we can generate the whole table

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