Additional Answer Options: 5.3.2

2023 - May/June - Paper 2 - Question 5

5.1

 

The initial concentration is 0 (of NO2)/concentration increases/ Curve starts at 0.check

Or

Curve B has an initial concentration and is the reactant as its concentration decreases.check

light bulb Tips:

  • The reaction that occurs is given by the equation: 2N2O5(g) → 4NO2(g)+O2(g).
  • Curve A starts from a concentration of zero therefore it must be a product because reactants have to react together for the concentration of the products to increase.
  • It is very unusual for a reaction to start in a sealed container where products are already present.

1 mark

 

5.2

 

Truecheck

n mol of N2O5 forms 2n mol of NO2 per unit time.check

Or

The gradient of the graph for NO2 is twice the gradient of the graph for N2O5.check

light bulbTip: the reaction that occurs is given by the equation: 2N2O5(g) → 4NO2(g)+O2(g). Therefore, the mol ratio of N2O5: NO2 is 2:4 which is simplified to 1:2.

2 marks

(2 x 1 mark)

5.3.1

 

light bulb How to approach this question:

  • Curve A gives the concentration of NO2 at 400s, the [NO2] is 200 x 10-4 mol·dm-3.
  • Do not forget to check the units of the axes as you can lose marks if you use a concentration of 200 mol·dm-3 instead of 200 x 10-4 mol·dm-3.
  • The molar mass of NO2 is 14 + 16 x 2 = 46g·mol-1, and the volume of the container is 2 dm3.

Option 1

\mathrm{c}\left(\mathrm{NO}_2\right)=\frac{\mathrm{m}}{\mathrm{MV}}check

200\times10^{-4}=\frac{m}{(46)(2)}checkcheck

\mathrm{m}=1,84\mathrm{~g}check

Or

Option 2

\mathrm{n}\left(\mathrm{NO}_2\right)=\mathrm{cV}check

=\left(200\times10^{-4}\right)\times2check

=4\times10^{-2}\mathrm{~mol}

\mathrm{n}\left(\mathrm{NO}_2\right)=\frac{\mathrm{m}}{\mathrm{M}}

4\times10^{-2}=\frac{m}{46}check

\mathrm{m}=1,84\mathrm{~g}check

4 marks

(4 x 1 mark)

Marking criteria:

  • Formula: c = m/MV or n(NO2) = cV / n(NO2) =m/Mcheck 
  • Substitute change in concentrationcheck 
  • Substitute M (46) and V (2)check 
  • The final correct answer: 1,84g.check

5.3.2

 

Option 1

\text{ Ave rate }=-\frac{\Delta\mathrm{c}\left(\mathrm{N}_2\mathrm{O}_5\right)}{\Delta\mathrm{t}}

=-\frac{\left(60\times10^{-4}-200\times10^{-4}\right)}{700(-0)}checkcheck

=2\times10^{-5}\left(\mathrm{~mol}\cdot\mathrm{dm}^{-3}\cdot\mathrm{s}^{-1}\right)

\text{ rate }\left(\mathrm{O}_2\right)=1/2\text{ rate }\left(\mathrm{N}_2\mathrm{O}_5\right)=1/2\left(2\times10^{-5}\right)check

=1\times10^{-5}\left(\mathrm{~mol}\cdot\mathrm{dm}^{-3}\cdot\mathrm{s}^{-1}\right)check

light bulb Tips:

  • The average rate of production of O2 cannot be determined directly from the graph since no curve shows the concentration of oxygen, therefore, you will need to work out the rate of change of NO2 or N2O5 and use the mol ratio.
  • If you use the rate of change of N2O5, you need to add a negative sign in the formula to make the rate of the reaction positive.
  • Always remember to look at the units for the rate of the reaction to work out the correct rate.

There are more ways to answer this question. To view other options, click here.

4 marks

(4 x 1 mark)

Marking criteria:

  • Substitute the change in concentration into the rate formulacheck 
  • Substitute time into the rate formulacheck
  • Use mol ratio: rate(O2) = ½ rate(N2O5)/ rate(O2) = ¼ rate(NO2).check 
  • Final correct answer: 1 x 10-5 (mol·dm-3·s-1)check

5.4.1

 

checkcheck 

 

light bulb Tips: 

  • It is essential that your diagram clearly shows both curves as the mark is given for curve Q having the same shape, a peak at the same kinetic energy, and being above curve P.
  • Both curves must start at the origin and the curves should not cross over each other.

2 marks

(2 x 1 mark)

Marking criteria:

  • Curve Q must be above the given curve P, and have the same shape as the given curve P, and the peaks have to correspondcheck
  • Start at the origin and do not cross curve Pcheck 

5.4.2

 

Higher thancheck

  • When the concentration of N2O5 is higher there are more N2O5 particles per unit volume.check
  • More effective collisions per unit time/second. /Higher frequency of effective collisions.check

light bulb Tip: increasing the concentration of reactant will increase the rate of the reaction. This is because there are more particles per unit volume and, therefore, more effective collisions per unit of time, leading to a higher rate of reaction.

3 marks

(3 x 1 mark)

Or

Option 2

\text{ Ave rate }=\frac{\Delta\mathrm{c}\left(\mathrm{NO}_2\right)}{\Delta\mathrm{t}}

=\frac{\left(280\times10^{-4}(-0)\right)}{700(-0)} (1 + 2)

=4\times10^{-5}\left(\mathrm{~mol}\cdot\mathrm{dm}^{-3}\cdot\mathrm{s}^{-1}\right)

\text{ rate }\left(\mathrm{O}_2\right)=1/4\operatorname{rate}\left(\mathrm{NO}_2\right)=1/4\left(4\times70^{-5}\right) (3)

=1\times10^{-5}\left(\mathrm{~mol}\cdot\mathrm{dm}^{-3}\cdot\mathrm{s}^{-1}\right) (4)

Or

Option 3

Part 1:

\Delta\mathrm{c}\left(\mathrm{O}_2\right)=1/2\Delta\mathrm{c}\left(\mathrm{N}_2\mathrm{O}_5\right)

=1/2\left(60\times10^{-4}-200\times10^{-4}\right) (1)

=1/2\left(140\times10^{-4}\right)

=7\times10^{-3}\mathrm{~mol}\cdot\mathrm{dm}^{-3}

Or

\Delta\mathrm{c}\left(\mathrm{O}_2\right)=1/4\Delta\mathrm{c}\left(\mathrm{NO}_2\right)

=1/4\left(280\times10^{-4}(-0)\right) (1)

=7\times10^{-3}\mathrm{~mol}\cdot\mathrm{dm}^{-3}

Part 2:

\text{ Ave rate }=\frac{\Delta\mathrm{c}\left(\mathrm{O}_2\right)}{\Delta\mathrm{t}}

=\frac{\left(7\times10^{-3}\right)}{700(-0)} (2 + 3)

=1\times10^{-5}\left(\mathrm{~mol}\cdot\mathrm{dm}^{-3}\cdot\mathrm{s}^{-1}\right) (4)

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