Question

The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. What is the rate immediately after the concentrations of all reactants are instantaneously lowered to 50% of their current value by adding an additional 50.0 mL of solvent?

Answer 1

0.0010 mol·L⁻¹s⁻¹

Step-by-step explanation:

Assume the rate law is

rate = k[A][B]²

If you are comparing two rates,

`\frac{\text{rate}_(2)}{\text{rate}_(1)} = \frac{k_(2)\text{[A]}_2[\text{B]}_(2)^(2)}{k_(1)\text{[A]}_1[\text{B]}_(1)^(2)}= \left (\frac{\text{[A]}_(2)}{\text{[A]}_(1)}\right ) \left (\frac{\text{[B]}_(2)}{\text{[B]}_(1)}\right )^(2)`

You are cutting each concentration in half, so

`\frac{\text{[A]}_(2)}{\text{[A]}_(1)} = (1)/(2)\text{ and }\frac{\text{[B]}_(2)}{\text{[B]}_(1)}= (1)/(2)`

Then,

`\frac{\text{rate}_(2)}{\text{rate}_(1)} = \left ((1)/(2)\right ) \left ((1)/(2)\right )^(2) = (1)/(2)*(1)/(4) = (1)/(8)\\\\\text{rate}_(2) = (1)/(8)* \text{rate}_(1)= (1)/(8)* \text{0.0080 mol$\cdot$L$^(-1)$s$^(-1)$} = \textbf{0.0010 mol$\cdot$L$^(-1)$s$^(-1)$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}`