How Fast? YR13 rate
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What is the general formula of the rate equation?
A + B —> C + D Rate = k[A]ᵐ[B]ⁿ Where m and n are the orders of reaction from 0 -2
What is the units for rate?
mol dm⁻³ s⁻¹
What is the equation to find the rate constant, k?
k = Rate/[A]ᵐ[B]ⁿ
What is meant by a zero order reaction?
The concentration of the substance has no impact on the rate Rate = k
What is meant by a 1st order reaction?
Doubling the concentration doubles the rate
What is meant by a 2nd order reaction?
rate = k[A]² Doubling the concentration will increase the rate 4x over Triping the rate will increase the concentration 9x over etc
What is the shape of a zero order rate-concentration graph?
A straight, horizontal line
What is the shape of a 1st order rate-concentration graph?
A straight line through the origin with increasing gradient
What is the shape of a 2nd order rate-concentration graph?
A steep curve upwards, starting from the origin
What is the shape of a zero order concentration-time graph?
A straight line with a decreasing gradient
What is the shape of a 1st order concentration-time graph?
A gently curved line with decreasing gradient, starting from the top left of the graph
What is the shape of a 2nd order concentration-time graph?
A steep curved line with decreasing gradient, starting from the top left of the graph
What is the definition of half life?
The time taken for the initial concentration of the reactants to decrease by half
How can a 1st order reaction be identified using half life?
A 1st order reaction will always have a constant half life
What is the equation to work out k when the half life of a reaction is known?
k = ln(2)/Half life
How do you prove that a steeply curving rate-concentration graph is definitively 2nd order?
• Square all concentration values • Keep rate values the same • Draw a graph with the new values. If the new graph is a straight line with a positive gradient, the original graph is 2nd order • K can then be calcualted by finding the gradient of the 2nd graph
What is the definition of rate determining step?
The slowest step in a multi step mechinism determines the rate
How do you write a rate equation when given the steps of a multi step reaction?
• Identify the reactants of the rate determining step • Add each reactant to the rate equation in [square brackets] • The number of molecules in the equation is the power
When given the rate equation, what is the process to deduce the steps of the equation?
Rate = k[NO₂]² Overall equation = NO₂ + CO -> NO + CO₂ * Look at what species are in the rate equation given. They are the reactants in the rate determining step = NO₂ + NO₂. 2 NO₂s are used because the rate equation had them to the power of 2. * Try to form a product in the overall equation and then balance as neccesary. NO₂ + NO₂.-> NO + NO₃. This is equation step 1. * Identify what species does not appear in overall equation but does in Step 1. This species must be cancelled out. * Try to react the desired cancellation species (NO₃) with a reactant from the overall equation that hasn’t been used yet. In this case, CO. NO₃ + CO -> CO₂ + NO₂ * If all reactant species have been used and all product species have been made, the steps should be correct.
What does each part of the Arrhenius Equation represent and what are the units (if applicable)?
• K = rate constant • A = pre-exponentional energy factor • e = natural log constant • -Ea = activation energy (J mol⁻¹) • R = Gas constant (given in data booklet) • T = Temprature (Kelvin)
What does the -Ea/RT portion of the Arrhenius equation tell us?
The proportion of particles that have exceeded Ea and therefore have sufficient energy for a reaction to take place
What does the pre-exponential energy factor, A, tell us?
The frequency of collisions with the correct orientation.
How is the pre-exponential energy factor, A, effected by temprarure?
It increases slightly but can be assumed to be constant over a small temprature change
What is the effect on the rate constant, k, when temprature increases? Explain in terms of the Maxwell-Boltzmann distrubution of molecular energies.
The temprature increase shifts the Boltzmann distribution to the right. The number of particles above the Ea increases, so they move faster and collide more frequently at the correct orientation.