Chemistry : Rates Of Reaction Help

[Rommarrooo]

O/N 11, Variant 41, Entire Q2, Help would be much appreciated. Thankyou.

 

[littlecloud11]

2ai) Comparing reaction 1 and 2 the conc. of CH3OH and H+ remains constant, but as the conc. CH3CHO increases by a factor of 1.25 the rate of reaction increases by a factor of 1.25. So the order with regard to CH3CHO is one.

Comparing reaction 2 and 3, the conc. of CH3OH increases by a factor of 1.6 and the conc. of the other reactants stay the same. The rate increases by a factor of 2/1.25= 1.6. So the order for CH3OH is also 1

comparing reaction 1 and 4, the conc. of CH3CHO remains constant and CH3OH and H+ are both varied. We already know the order for CH3OH is 1 so we can eliminate the increase in rate caused by this. As the conc of CH3OH increases by 1.6 the rate is also supposed to increase by 1.6. The rate between reaction 1 and 4 increases by 3.2, the increase due to CH3OH is by a factor of 1.6, so the increase due to H+ is 3.2/1.6= 2
You can see that as the conc of H+ is doubled the rate doubles, so the order for H+ is also 1

ii) rate eq =[CH3OH]x [CH3CHO]x[H+]

iii) the unit would be (mol dm^-3 s^-1) / (mol^3 dm^-9) = mol^-2 dm^6 s^-1

iv) For finding the relative rate use any of the reaction in the chart and find the rate when the conc. is .2 for each reactant. Like for reaction 1-
the conc. of CH3CHO is already .2, The conc. for CH3OH has to be doubled so the rate has to be doubled. The conc. for H+ has to be increased by a factor of .2/.05 = 4 so the rate would increase 4 times.
The overall increase would therefore 2*4 = 8 times that of reaction 1, which is 8

bi) Use the reaction given in the question. One mole ethanal reaction with 2 mols of methanol. So as x moles of ethanal is used up 2x moles of methanol is used. So the equilibrium conc. for methanol would be (.1-2x). H+ is the catalyst, it's conc. remains the same throughout the reaction. One mole of ethanal produces one mole of acetate and 1 mole of H2O, 'x' moles of ethanal was used so the equilibrium conc. for each product would be x.

bii) you can see that the value of x is .025 as the conc for acetate is equal to x. The conc. of H2O would also be x i.e. .025
The conc. for ethanal would be .2-.025 = .175
The conc. for Methanol would be .1-.025*2 =.05

biii) Kc = [acetate]x[H2o]/ [Ethanal]x [Methanol]^2

iv) Kc = .025*.025/ (.175) * (.05)^2 = 1.43

 

[Rommarrooo]

Thankyou very much.