The Chemical Industry

    Note: specification points g and i are not on this page because they require no knowledge and just require thought in the exam.

a Rate of reaction:

  • Rate of reaction measures the rate of reactantproduct conversion. It can easily be calculated from the gradient of a concentration-time graph
  • Rate equations can be used to calculate the rate of reaction from concentrations and a known constant for a given temperature
  • rate = k[A][B]... where:
    - k is the rate constant, which is different for each reaction and temperature
    - [X] represents the concentration of a reactant/catalyst
    - etc represents the order of that reactant/catalyst (see below)
  • As temperature increases, the rate constant k increases


  • Order is an integer unique to each reactant, which can be determined by experimental data only
  • Overall order is the sum of all individual orders in a reaction
  • Orders vary for each reaction and are calculated by measuring rate of reaction increase when the concentration of that reactant doubles
  • When finding the order of a reactant, ensure that all other reactants are in excess
  • For example, if doubling concentration for reactant A doubles rate of reaction, it is linear - the order is 1
  • If doubling concentration for a reactant quadruples rate of reaction, it is a quadratic relationship with order 2
  • This can be written with the proportionality symbol ∝
    - With order 1 for reactant A, rate [A]
    - If it was order 2, it would be rate [A]
  • Alternatively, the ∝ symbol could be removed and the equations written like rate = k[A] or rate = k[A]

b Reaction half-lives:

  • Reaction half-life is the time it takes for half of a reactant to be used up
  • A half-life can be easily calculated from a concentration-time graph by finding the time taken for the concentration to halve
  • By measuring 2-3 half-lives, the order can be determined:
    - If half-life decreases with time, it's zero order
    - If half-life remains constant, it's first order
    - If half-life increases with time, it's second order
  • The symbol for a half-life is

Calculating rate constant from half-life (first order only):

  • The half-life remains constant for a first-order reaction, so the rate constant can be calculated with (units are )
  • The derivation of this equation is not required

c Initial rate:

  • The initial rate of reaction is the rate at the start of the reaction. It is calculated by measuring the time taken for a small amount of a reactant to be used or a small amount of product to form, and using the equation initial rate = amount of reactant used/product formed time
  • It can also be estimated from the gradient at the start of a concentration-time graph

Clock reactions:

  • A clock reaction has a clear end point (e.g. a colour change) once a certain amount of product has formed. Therefore, clock reactions are useful for measuring initial rate of reaction
    TO DO: Cards from here
  • The iodine clock reaction is H2O2 (hydrogen peroxide) + 2I- + 2H+ 2H2O + I2. Sodium thiosulfate instantly removes iodine, so a small amount is added. Once this is all used up, there will be iodine molecules in the solution, which can be tested for with starch (which turns blue/black in the presence of iodine)

Concentration-time graphs:

  • On a concentration-time graph, if the reaction is zero-order (with respect to the reactant being investigated), it is a negative-gradient straight line, because the reactant concentration decreases at a constant rate
  • If first order, the graph has an exponential decay
    TO DO: or is it quadratic?
    curve, because rate increases with reactant concentration, so rate decreases over time as the reactant is used up
  • If second order, the graph has a similar shape, but steeper gradient, as the rate decrease per second is increasing over the course of the reaction

Rate-concentration graphs:

  • For a zero-order reactant, this is a horizontal line - the rate remains constant for any concentration
  • For a first-order reactant, the rate-concentration graph has a straight line with positive gradient - rate is increasing with concentration
  • Second order reactants will have a quadratic graph (or a straight line when rate is plotted against concentration)

d The Arrhenius equation:

  • The Arrhenius equation is k = Ae (given in data sheet), where:
    - k is the rate constant
    - is the activation enthalpy (J mol-1)
    - T is the temperature (K)
    - R is the gas constant (8.314, given in data sheet)
    - A is the pre-exponential factor
  • Taking logs of both sides: ln k = + ln A (also given in data sheet)
  • This is in y = mx + c form, so ln k can be plotted against . The resulting graph will have a gradient of and y-intercept of ln A
  • Remember that the calculated activation enthalpy will be in J mol-1 and temperature will be in kelvins, so remember to convert if required

e Rate-determining steps:

  • In a multi-step reaction, the rate-determining step is the step with the lowest rate. This step determines the order of the reaction
  • If a reactant isn't in the rate equation, it can't be in the rate-determining step
  • Catalysts can be in rate-determining steps (they can appear in rate equations sometimes)
  • If given the equation of the rate-determining step, the order of a reactant can be predicted from the number of times it appears
  • However, this method doesn't always work - for example with the reaction 2A + B C it would be assumed that the reaction is second order with respect to A, but it could also be first-order if only one A is present in the rate determining step and the second A is involved in another, much quicker, step

f Factors affecting the equilibrium constant:

  • Increasing the pressure of a dynamic equilibrium will move the position of the equilibrium towards the side with the fewest moles of gas. It will not affect
  • Changing the concentration of a reactant or product in an equilibrium will move the position, but also not affect the value of
  • Adding a catalyst will not affect the position of equilibrium or value of , but they do reduce the time taken for the equilibrium to form

  • The three values above don't affect because the proportions of reactants and products change in a way to keep it the same
  • However, changing the temperature of an equilibrium will affect both the position of equilibrium and value of

h Equilibrium calculations:

  • The fraction used to find can be rearranged to find unknowns

Finding an equilibrium constant experimentally:

  • If one of the substances is coloured, colorimetry can be used, with the calibration curve being used to calculate the concentration of the coloured substance
  • If one of the sides of the reaction contains an acid or alkali, a pH probe can be used. If the equilibrium took a long time to form, use a titration
  • But do not use a titration for equilibria with a high speed of formation because adding an acid/alkali would significantly change the position of equilibrium, giving a different, inaccurate value for concentration

j Nitrogen chemistry:

  • Nitrogen atoms bond diatomically to form N2 molecules with a triple bond, therefore giving it a very high bond enthalpy
  • Ammonia has the structure NH3 with a lone pair
  • An ammonium ion is NH4+

Nitrogen oxides:

  • NO is nitrogen monoxide / nitrogen(II) oxide, a colourless gas
  • N2O is dinitrogen monoxide / nitrogen(I) oxide, a colourless gas with a sweet smell (laughing gas)
  • NO2 is nitrogen dioxide / nitrogen(IV) oxide, a brown gas with sharp odour

Testing for ammonium compounds:

  • Add NaOH and heat. Ammonium ions will react with the OH- ions, releasing ammonia gas (which will turn damp red litmus paper blue)

Testing for nitrate(V) ions:

  • Nitrate(V) ions have the formula NO3-
  • To test for them, heat the solution with NaOH and add a piece of Devarda's alloy (mainly aluminium and copper)
  • The aluminium reduces the nitrate(V) ions, creating ammonia gas, water and Al(OH)4-

The nitrogen cycle:

  • N2 and H2 react to form ammonia
  • (ammonia) + H+ creates ammonium ions
  • (ammonium ions) + O2 form nitrate(III) ions (NO2-)
  • (nitrate(III) ions) + H2O creates nitrate(V) ions (NO3-)
  • (nitrate(V) ions) + H+ + e- forms nitrogen gas again

  • Oxides are formed with N2 and a multiple of O2