rate equations [P2] PAPER 2 Flashcards
general rate equation?
rate = k [ ….]
• doesn’t appear in rate equation if zero order in respect to [ ]
The rate equation for a reaction is rate = k[E].
Explain qualitatively why doubling the temperature has a much greater effect on the rate of the reaction than doubling the concentration of E.
• reaction occurs when molecules have E ≥ Ea
• doubling temperature causes MANY more molecules to have this Ea
• Whereas doubling [E] only doubles the number of molecules with this Ea
Arrhenius equation?
The rate of reaction is first order with respect to A. If the concentration of A is doubled, what happens to the rate?
rate doubles
The rate of reaction is second order with respect to A. If the concentration of A is doubled, what happens to the rate?
rate is multiplied by 2^2 (so x4)
The rate of reaction is zero order with respect to A. If the concentration of A is doubled, what happens to the rate?
nothing
What factors affect the value of the rate constant?
temperature
what is meant by the term ‘order of reaction’ with respect to __?
power of concentration term in rate equation
How to deduce the overall order of a reaction?
add all the powers in the rate equation together
How to determine the rate-determining step?
the step at which all the species involved in the rate equation are used up by
The rate equation for a reaction between A and B is
rate = k[A].
The reaction is zero order with respect to B. State the significance of this zero order for the mechanism of the reaction.
rate-determining step
involves only A
Explain why step __ is the rate-determining step
species in step __ are in proportion as in the rate equation
State the effect, if any, on the value of the rate constant k when the temperature is lowered but all other conditions are kept constant. Explain your answer.
• value of k is lower
• fewer particles have energy > Ea
In a further experiment at a different temperature, the initial rate of reaction was found to be 9.0 × 10–3 mol dm–3 s–1 when the initial concentration of A was 0.020 mol dm–3 and the initial concentration of NaOH was 2.00 mol dm–3. Under these new conditions with the much higher concentration of sodium hydroxide, the reaction is first order with respect to A and appears to be zero order with respect to sodium hydroxide.
Suggest why the order of reaction with respect to sodium hydroxide appears to be zero under these new conditions.
• large excess of OH–
• [OH–] is effectively constant, so has no effect on the rate
How can you tell that H+ acts as a catalyst in this reaction?
appears in the rate equation, but doesn’t appear in the overall equation
rate-concentration graph to show zero order with respect to [ ]
rate-concentration graph to show first order with respect to [ ]
rate-concentration graph to show second order with respect to [ ]
constant gradient as [H+] decreases
• measure known volumes of A, B, C and X into separate beakers
• mix A, B and X together and place in a water bath to ensure temperature is constant
• add C to this mixture, start a timer at the point of mixing, and record the time it takes for the mixture to turn blue
• repeat with different concentrations of A (by changing the volumes of A but keeping the volume of the overall solution the same)
• 1/time taken is a measure of the rate
• plot a graph of 1/time against concentrations of A
• if the graph is horizontal, then order of reaction is zero with respect to A, if a straight line through the origin, then it is first order, and if it is curved and through the origin, then it is second order
Explain how the graph shows that the reaction is zero-order with respect to iodine in the reaction between propanone and iodine.
The graph has a constant gradient, so the iodine is being used up at a constant rate
• plot concentration (y-axis) against time (x-axis), and take tangents to find the gradients, so calculate rates
• plot rates against concentration
• straight line through the origin (directly proportional) confirms first order
Explain how the graph shows that the reaction is zero-order with respect to iodine in the reaction between propanone and iodine.
• graph has a constant gradient
• so, the rate of reaction doesn’t change as the concentration of iodine changes (the iodine is being used up at a constant rate)
