# 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 reactant→product 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

Test yourself

## Order:

**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

Test yourself

## 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

Test yourself

## 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
`H`

. 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)_{2}O_{2}(hydrogen peroxide) + 2I^{-}+ 2H^{+}→ 2H_{2}O + I_{2}

## 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 curve, because rate increases with reactant concentration, so rate decreases over time as the reactant is used up
**TO DO: or is it quadratic?** - 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

Test yourself

**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

Test yourself

**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

Test yourself

**h** Equilibrium calculations:

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

Test yourself

## 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 N
_{2}molecules with a triple bond, therefore giving it a very high bond enthalpy - Ammonia has the structure NH
_{3}with a lone pair - An ammonium ion is NH
_{4}^{+}

Test yourself

## Nitrogen oxides:

- NO is
*nitrogen monoxide*/*nitrogen(II) oxide*, a colourless gas - N
_{2}O is*dinitrogen monoxide*/*nitrogen(I) oxide*, a colourless gas with a sweet smell (*laughing gas*) - NO
_{2}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 NO
_{3}^{-} - 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:

- N
_{2}and H_{2}react to form ammonia - (ammonia) + H
^{+}creates ammonium ions - (ammonium ions) + O
_{2}form nitrate(III) ions (NO_{2}^{-}) - (nitrate(III) ions) + H
_{2}O creates nitrate(V) ions (NO_{3}^{-}) - (nitrate(V) ions) + H
^{+}+ e^{-}forms nitrogen gas again - Oxides are formed with N
_{2}and a multiple of O_{2}