Rates 2 Flashcards
Order of reaction
The way in which a conc of reactant affects the rate
Zero order
Rate = [A]^0
Conc of reactant has no effect on the rate
First order
When the rate depends on its conc raised to the power of one
If conc A triples rate also triples
Second order
When the rate depends on its concentration raised to the power of 2
If conc A triples rate x9
Rate equation
Rate = k[A]^m[B]^n
Overall order = M+N
Conc time graphs
Zero order - straight line negative gradient gradient = k
First order - downward curve
k=ln2/half life
Second order - downward curve steeper at the start tailing off more slowly
Rate conc graphs
Zero order - horizontal line
Rate=k
First order - k[A] straight line through origin proportional
Second order - rate=k[A]^2 curve
Plot again conc^2 straight line through origin
Initial rate
instantaneous rate at start of reaction
Can be found measuring gradient of tangent at t=0 on conc time graph
Initial rate of clock = 1/t
Rate determining step
Slowest step in a reaction
Multiple step reactions
Eg rate = k [x]^2
2X -> Xy+ Xc
Xc +Y -> 2XY
Factors affecting the rate constant
Increasing the temp increases the activation energy
Arrhenius equation
K= A x e^-Ea/RT
A= (pre exponential factor)
R= Gas constant
T = temp
Arrhenius plot
ln k = -Ea/R x 1/t + lnA
-Ea/R = gradient
1/t= x value
lnA = y intercept
Equilibrium constant Kc
Only gases
If homogeneous species Kc uses all in equation
If heterozygous species Kc only uses gases in the equation
Calculating equilibrium quant to calculate Kc
Reacting amounts
Initial
Change
Equilibrium
USE RICE TABLES
Divide by total volume and sun into to Kc equation
Kp
Is the equilibrium constant used for gases
Mole fraction
No of moles of A/No of moles in gas mixture
Partial pressure
Contribution has makes towards the total pressure
Mole fraction x total pressure
Kp equation
p(products)/p(reactants)
K=1
Equilibrium half way between products and reactants
K=100
Equilibrium well in favour of products
K=1x10^-3
Equilibrium well in favour of reactants
Exothermic effects on K
If forwards reaction is Exothermic
Raising the temp decreases K
Due to equilibrium shifting more towards reactants
Explaining equil shift
System is no longer in equilibrium and Kp ratio is greater than Kp
Equil partial pressures must change
Product decr reactants incr
Equil shifts to the left
How does conc affect K
If conc of reactants increased the ratio is now less than Kc so no longer in equil
Conc of product incr conc of reactant decr
New equilibrium shifts right to restore Kc value
How does pressure affect K
If reactant pressure incr and product
Product must decrease and reactant increase to restore Kp value
