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Summary AQA A Level Chemistry Rate Equations Notes
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AQA A Level Chemistry Rate Equations Summarised Notes. Made these on iPad from textbook, youtube videos and exam questions.
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RATE EQUATIONSCALCULATING RATE FROM A GRAPH
in
Rate can be found from the gradient orange
a
cuse a
bigger section of the
graph
Rate can also be found from the of a tangent
foot on a curved line of best fit.
↳ With curves, draw a
tangent
across the
4 extend the line right graph
4 the gradient is the rate at the specific point on the curve.
y
"a tangent
THE RATE EQUATION ORDERS
The rate equation links rate with entrations of substances
rate =
k(AT"(B]*
conc. of substance
rate/moldmn-s- -
e.g. in
following
equ. A + B -> c+ D
rate constant
units vary
ORDERS OF REACTION
↳ an order is the power to which
- -
a conc. is raised to in the rate equation. It tells us how the cone of
the substance affects the rate.
·
ICROORDER: changes in conc. that have no effect on rate
e.g.
if (A] doubles then rate doesn't change.
·PORACR: changes in conc has a proportional change on rate e.g. if [A] doubles then rate
doubles.
,·
INPORDER: changes in conc. has a
squared proportional change on a rate e.g. if [A] doubles then rate
quadruples.
!orders can
only be determined by an experiment, you can't work it out by looking at an equation.
The Rate Equation Rate a Constant
The rate equ. Links rate with cone of substances.
rate constant K -
a constant no that allows us to ->
only fixed at a particular temperature,
equate rate a conc. if the temperature changes so does the
rate constant. ('K'increases when temp
increases).
- the larger the value of is, the faster the rate of reaction.
↳
we increase the temperature, the panicles have kinetic collide often, this increases
as more
energy a more
the rate
rate k/A]"/B]*
=
↳
but the cone of the substances remain constant.
↳to make this
equ balance the value of multicoreace.
CALCULATING RATE
be calculated the rate equation.
rate can
by using
·
EXAMPLE: The following reaction was carried out in a lab.
2
NO,g +
O,(g) ->
2NOc(g)
It was found that the reaction is "order with respect to NO and I order with respect to 02. Calculate
the rate of reaction given the conc. of NO 3x10% and 0, is 1x10%. The rate constant in
700,000 mol "dm's-'
↳ 1: write rate expression
rate k(No] [0,]'
=
↳ 2: substitute the numbers
rate= 700,000 x (3x10%'x((x10) =
6.3x10 "moldmn"s
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