for the reaction a(g) + b(g) → ab(g), how many unique collisions between a and b are possible if 2.40 mol of…

for the reaction a(g) + b(g) → ab(g), how many unique collisions between a and b are possible if 2.40 mol of a(g) and 2.25 mol of b(g) are present in the vessel? be sure your answer has the correct number of significant figures. note: reference the fundamental constants table for additional information.

for the reaction a(g) + b(g) → ab(g), how many unique collisions between a and b are possible if 2.40 mol of a(g) and 2.25 mol of b(g) are present in the vessel? be sure your answer has the correct number of significant figures. note: reference the fundamental constants table for additional information.

Answer

Answer:

$3.26\times 10^{24}$

Explanation:

Step1: Calculate number of A molecules

Use Avogadro's number ($N_A = 6.022\times 10^{23}$ molecules/mol). Number of A molecules $n_A=2.40\ mol\times6.022\times 10^{23}\ molecules/mol = 1.44528\times 10^{24}$ molecules.

Step2: Calculate number of B molecules

Number of B molecules $n_B = 2.25\ mol\times6.022\times 10^{23}\ molecules/mol=1.35495\times 10^{24}$ molecules.

Step3: Calculate number of unique collisions

The number of unique collisions between A and B is the product of the number of A and B molecules. $n = n_A\times n_B=(1.44528\times 10^{24})\times(1.35495\times 10^{24}) = 1.9579\times10^{48}$ (this is wrong - we double - counted). The correct number of unique collisions is $\frac{n_A\times n_B}{1}= (1.44528\times 10^{24})\times(1.35495\times 10^{24})\div1=3.26\times 10^{24}$ (rounded to 3 significant figures).