Question

For the reaction $\mathbf{A}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{B}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{A}\mathbf{B}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$, 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

Step 1: 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 A:

$$\begin{gathered} ThenumberofAparticles=1.01molA\times \frac{{6.022\times 10}^{23}A}{molA} \\ =6.08222\times {10}^{23}A \end{gathered}$$

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

$$\begin{gathered} ThenumberofBparticles=2.12molB\times \frac{{6.022\times 10}^{23}B}{molB} \\ =12.76664\times {10}^{23}B \end{gathered}$$

Step 2: 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:

$$\begin{gathered} Thenumberofuniquecollision=\left({6.08222\times 10}^{23}\right)\times \left({12.76664\times 10}^{23}\right) \\ =7.46\times {10}^{47} \end{gathered}$$

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