Rate of equations

r = k[A]m[B]n m, n are called reaction orders Orders are usually integers 0,1,2 0 means the reaction is zero order with respect to that reactant 1 means first order 2 means second-order NOTE: the orders are not the same as the stoichiometric coefficients in the balanced equation. They are worked out experimentally.

The total order for a reaction is worked out by adding all the individual orders together (m+n)

• The units of k depend on the overall order of the reaction. It must be worked out from the rate equation • The value of k is independent of concentration and time. It is constant at a fixed temperature • The value of k refers to a specific temperature and it increases if we increase temperature

When we follow one experiment over time recording the change in concentration we call it a continuous rate method. The gradient represents the rate of reaction. The reaction is fastest at the start where the gradient is steepest. The rate drops as the reactants start to get used up and their concentration drops. The graph will eventually become horizontal and the gradient becomes zero which represents the reaction having stopped.

In reactions where there are several reactants, if the concentration of one of the reactant is kept in a large excess then that reactant will appear not to affect rate and will be pseudo-zero order . This is because its concentration stays virtually constant and does not affect rate.

The initial rate can be calculated from taking the gradient of a continuous monitoring conc vs time graph at time = zero Initial rate can also be calculated from clock reactions where the time taken to reach a fixed concentration is measured.

For zero order reactants, the rate stays constant as the reactant is used up. This means the concentration of that reactant has no effect on rate. Rate = k [A]0 so rate = k As the rate is the gradient of the graph on the right, the gradient is also the value of the rate constant

Increasing the temperature increases the value of the rate constant k . The relationship is given by the Arrhenius equation k = Ae-EA/RT where A is the Arrhenius constant, R is the gas constant, and EA is the activation energy. ln k = ln A – EA/(RT)

• The concentration of the reactant does not change significantly over the timescale • The temperature is constant • The endpoint has not been over-estimated by observing the colour change or precipitate formed too late

• Precipitate formation — if the solution is clear and the reaction produces a precipitate which makes it opaque, then the time taken for us to no longer be able to see through the solution is measured • Colour change — colorimetry is a useful technique if there is a colour change, although if the colour change is very distinct, colorimetry is not needed

• H+ ions and hydrogen peroxide oxidise I– ions to form I 2 H2O2 (aq) + 2I– (aq) + 2H+ (aq)  I2 (aq) + 2H2O (l) • A small amount of sodium thiosulphate with a starch indicator is added to give an orange solution 2S2O3 2– (aq) + I2 (aq)  2I– (aq) + S4O6 2– (aq) Any iodine produced by the first reaction immediately reacts with thiosulphate ions, converting them back to I - ions Eventually, sodium thiosulphate will be used up and I 2 will accumulate, causing the starch to change colour from orange to dark blue at the endpoint By repeating this reaction with different concentrations of one of the reagents and plotting a graph of rate against concentration, the order of reaction with respect to this reagent can be determined

The partial pressure of a gas in a mixture is the pressure that the gas would have if it alone occupied the volume occupied by the whole mixture. If a mixture of gases contains 3 different gases then the total pressure will equal the 3 partial pressure added together P =p1 + p2 + p3

partial pressure = mole fraction x total pressure

mole fraction = number of moles of a gas total number of moles of all gases

The larger the Kp the greater the number of products. If Kp is small we say the equilibrium favors the reactants Kc and Kp only change with temperature. It does not change if pressure or concentration is altered. A catalyst also has no effect on Kc or Kp

Both the position of equilibrium and the value of Kc or Kp will change it temperature is altered In this equilibrium which is exothermic in the forward direction If temperature is increased the reaction will shift to oppose the change and move in the backwards endothermic direction. The position of equilibrium shifts left. The value of Kp gets smaller as there are fewer products