If the rate of a reaction increases by a factor of 64 when the concentration of reactant increases by a factor of 4, what is the order of the reaction with respect to this reactant?

The reaction is of order three with respect to the reactant.

The rate of a reaction of order n about a certain reactant is proportion to the concentration of that reactant raised to the n-th power. This is true only if  concentrations of any other reactants stay constant in the whole process.

In other words, Rate = constant × [Reactant]ⁿ, Rate ∝ [Reactant]ⁿ. (The symbol "∝" reads "proportional to".)

In this question,

[4 × Reactant]ⁿ ÷ [Reactant]ⁿ = 64.

In other words, 4ⁿ = 64, where n is the order of the reaction with respect to this reactant.

It might take some guesswork to find the value of n. Alternatively, n can be solved directly with a calculator using logarithms. Taking natural log of both sides:

.

Evaluating on Google or on a calculator with support for ln (the natural log) will give the value of n- no guesswork required.

n = 3. Therefore, the reaction is of order three with respect to this reactant.

For A: The average molecular speed of Ne gas is 553 m/s at the same temperature.

For B: The rate of effusion of gas is

Explanation:

For A:

The average molecular speed of the gas is calculated by using the formula:

     OR

where, M is the molar mass of gas

Forming an equation for the two gases:

         .....(1)

Given values:

Plugging values in equation 1:

Hence, the average molecular speed of Ne gas is 553 m/s at the same temperature.

For B:

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation for this follows:

Where, M is the molar mass of the gas

Forming an equation for the two gases:

         .....(2)

Given values:

Plugging values in equation 2:

Hence, the rate of effusion of gas is