Question

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

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

Step-by-step solution

The number of A and B particles

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

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

$$\begin{gathered} \mathrm{The}\mathrm{number}\mathrm{of}\mathrm{B}\mathrm{particles}=2.12\mathrm{mol}\mathrm{B}\times \frac{6.022\times {10}^{23}\mathrm{B}}{\mathrm{mol}\mathrm{B}} \\ =12.76664\times {10}^{23}\mathrm{B} \end{gathered}$$

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:

$$\begin{gathered} \mathrm{The}\mathrm{number}\mathrm{of}\mathrm{unique}\mathrm{collision}=\left(6.08222\times {10}^{23}\right)\times \left(12.76664\times {10}^{23}\right) \\ =7.46\times {10}^{47} \end{gathered}$$

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