Rates of Reaction

Flask A = 1g Zn, 100cm³ 1moldm^-3 HCl at 35C

Flask B = 1g Zn, 100cm³ 2moldm^-3 HCl at 35C

Explain why the rate of reaction in Flasks A and B are different?

The concentration of the acid is higher in Flask B, so the frequency of collisions of acid particles with zinc is higher, resulting in a faster rate of reaction.

Conc-Time graph for 1st Order Reactant

Conc-Time graph for 2nd Order Reactant

Conc-Time graph for 0th Order Reactant

Rate-Conc graph for 1st Order Reactant

Rate-Conc graph for 2nd Order Reactant

Rate-Conc graph for 0th Order Reactant

Arrhenius Equation

What does each letter in the Arrhenius equation stand for?

k = rate constant

A = pre-exponential factor (frequency factor)

e = base number of natural logarithms (SHIFT-ln on calculator)

Ea = Activation energy ( J mol⁻¹ )

R = Gas constant (8.314 J mol⁻¹ K⁻¹ )

T = Temperature in kelvin

What's the effect on the value of k when the temperature is increased?

Increases k

More frequent collisions and more collisions have energy equal to or greater than the activation energy.

What's the effect on the value of k when the activation energy is increased?

Decreases k

Fewer collisions have energy equal to or greater than the activation energy.

What's the effect on the value of k when the pre-exponential factor is decreased?

Less frequent collisions

Logarithmic Form of Arrhenius Equation

Compare the log form of Arrhenius Equation with a straight line:

What does the gradient of the graph (m) represent?

What does the y-intercept of the graph (c) represent?

How to work out Ea from the graph?

Gradient = -Ea/RT

Ea = -Gradient * R

How to calculate A?

lnk = -Ea/RT + lnA

lnA = lnk + Ea/RT

A = e^lnA

Ea the subject

Ea = RT (lnA - lnk)

A the subject

A = k * e^Ea/RT

T the subject

Use the results from the student’s experiments to deduce the rate equation for this reaction. Justify your reasoning.

Straight line through origin 1st order wrt BrO₃^-

Between expt 1 and 2, [H⁺] x 2, rate x 4

Rate ∝ [H⁺] ² , so 2nd order wrt H⁺

When [H⁺] constant and [BrO₃^-] x2 and [Br^-] x 2, rate x 4

Since BrO₃ ^- is first order, doubling [Br^- ] has doubled rate, so it is 1st order wrt Br^-

Rate = k [BrO₃^-][Br^-][H⁺]²