Topic 7

Question 1

state a change that can be made to apparatus to measure the volume of gas more accurately

Answer 1

use a burette

Question 2

a tangent has been drawn on the line on the graph. calculate the rate of reaction at this point.

Answer 2

difference in volume / difference in time

= 43-15 / 60-0
= 0.47

Question 3

two substances, A and B, each form a colourless solution. if the solutions are mixed in a beaker, A and B react to form a coloured product. the rate of reaction between A and B can be investigated by placing the beaker containing mixture on a cross on a piece of paper and timing how long it takes for enough coloured product to be produced to make the cross invisible when viewed from above, through the solution.

concentration of A in solution in g dm-3) 1 - 10, 2 - 10, 3 - 40
temperature in degrees) 1 - 20, 2 - 40, 3 - 40
time for cross to become invisible in s) 1 - 320, 2 - 80, 3 - 20

use these results to explain, in terms of the behaviour of particles, the effect of changing temperature and concentration of A in solution on the rate of reaction.

Answer 3

reactions occur when particle collisions have sufficient energy. reaction rates are increased when the collision frequencies are increased.

experiment 2 was carried out at a higher temperature than 1 so the particles have more kinetic energy so move faster, so there are more frequent collisions between particles in solutions A and B.

experiment 3 was carried out at a higher concentration than 2 so there are more reacting particles in the same volume of mixture therefore increasing frequency of successful collisions.

concentrations of reactants are the same in 1 and 2. reaction rate in 3 is the greatest due to combined effects of increased temperature and concentration

Question 4

hydrogen reacts with oxygen to form steam

2H2(g) + O2(g) —> 2H2O(g)

bond energies:
H—H - 435
O=O - 500
O—H - 460

calculate the energy change for the reaction of 2 mol of hydrogen gas H2, with 1 mol of oxygen gas O2, to give 2 mol of steam, H2O

Answer 4

broken = (2x435) + 500 = 1370

made = (4x460) = 1840

change = 1370 - 1840 = -470

exothermic

= -470

Question 5

explain in terms of bond breaking and bond making, why some reactions are exothermic

Answer 5

breaking bonds needs energy (endothermic) and making bonds releases energy (exothermic). more energy is taken in than given out to surroundings

Question 6

calcium carbonate reacts with dilute hydrochloric acid to produce carbon dioxide gas. the rate of reaction between calcium carbonate and dilute hydrochloric acid at room temp was investigated.
the investigation was carried out with different sized calcium carbonate pieces.

large - volume of carbon dioxide = 16

small - volume of carbon dioxide = 48

powder - volume of carbon dioxide = 90

state using the info the effect of surface area of the calcium carbonate on this rate of reaction

Answer 6

larger surface area means a faster rate

Question 7

calcium carbonate powder produced 90cm3 of carbon dioxide in five minutes. calculate the average rate of reaction in cm3s-1

Answer 7

5 x 60 = 300

rate = v/t
= 90 / 300

= 0.3

Question 8

the experiments were repeated at a higher temp. the rate of reaction for each experiment increased.

explain in terms of particles, why the rate of reaction increased when the temperature was increased

Answer 8

particles have more energy so move faster so there are more frequent collisions between particles and collide with sufficient energy

Question 9

sodium thiosulfate solution reacts with dilute hydrochloric acid. when dilute hydrochloric acid is mixed with sodium thiosulfate solution, the mixture turns cloudy. explain why

Answer 9

a solid precipitate of sulfur forms

Question 10

in an investigation, different concentrations of hydrochloric acid are reacted with sodium thiosulfate solution. the mixture goes cloudy at different rates.

describe how the rate at which the mixture goes cloudy can be measured

Answer 10

flask placed in front of cross then you measure time when cross is obscured

Question 11

you are provided with some dilute hydrochloric acid which has a concentration of 50 g dm-3. for this experiment dilute hydrochloric acid with a concentration of 20 g dm-3 is required. how much water must be added to 100cm3 of 50 g dm-3 hydrochloric acid to make dilute hydrochloric acid with a concentration of 20 g dm-3

A) 200 cm3
B) 150 cm3
C) 100 cm3
D) 50 cm3

Answer 11

B - 150 cm3

Question 12

C—H = 435
O=O = 496
C=O = 805
H—O = 463

calculate the energy change for:

CH4 + 2O2 —> CO2 + 2OH2

Answer 12

broken = (4x435) + (2x496) = 2732

made = (2x805) + (4x463) = 3462

change = broken - made = 2732 - 3462

= -730