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Chapter 16: Q16.59P (page 731)

For the reaction , how many unique collisions between A and B are possible if 1.01 mol of A(g) and 2.12 mol of B(g) are present in the vessel?

Short Answer

Expert verified

There are $7.46 脳 10^{47}$ possible collisions between particles A and B.

Step by step solution

01

The number of A and B particles

A material has $6.022 脳 10^{23}$ particles per mole. Then, as follows, calculate the number of particles in 1.01 moles $The number of A particles = 1.01 mol A 脳 \frac{6.022 脳 10^{23} A}{mol A} = 6.08222 脳 10^{23} A$

Determine the number of particles in 2.12 moles B in the same way:

$The number of B particles = 2.12 mol B 脳 \frac{6.022 脳 10^{23} B}{mol B} = 12.76664 脳 10^{23} B$

02

The number of unique collision between A and B

To find the number of distinct collisions between A and B, multiply the number of particles of each:

$The number of unique collision = (6.08222 脳 10^{23}) 脳 (12.76664 脳 10^{23}) = 7.46 脳 10^{47}$

As a result, there are $7.46 脳 10^{47}$ possible collisions between particles A and B.

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The growth of Pseudomonas bacteria is modelled as a first-order process with k = 0.035 $m i n^{- 1}$ at 37掳C. The initial Pseudomonas population density is $1.0 脳 10^{- 3}$ cells/L.

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$2 N_{5 (g)} 鈫� 4 {NO}_{2 (g)} + O_{2 (g),}$ the rate constant is k $= 2.8 脳 10^{- 3} s^{- 1} at 60 掳 C . The$ initial concentration of $N_{2} O_{5}$ is 1.58 mol/L.

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