K.C.S.E Chemistry Q & A - MODEL 2007.PP2.QN.4

14/10/2020

KCSE Chemistry 2007 Topical Questions and Answers

All Questions Featuring Chemistry Form 1, Form 2, Form 3, Form 4, Paper 1, Paper 2 and Paper 3

(a) Methanol is manufactured from carbon (IV) oxide and hydrogen gas according to the equation:

The reaction is carried out in the presence of a chromium catalyst at 700K and 30kPa. Under these conditions, equilibrium is reacted when 2% of the carbon (IV) oxide is converted to methanol
(i) How does the rate of the forward reaction compare with that of the reverse reaction when 2% of the carbon (IV) oxide is converted to methanol?
(ii) Explain how each of the following would affect the yield of methanol:
I Reduction
II Using a more efficient catalyst
(iii) If the reaction is carried out at 500K and 30kPa, the percentage of carbon (IV) oxide converted to methanol is higher than 2%
I what is the sign of ΔH for the reaction? Give a reason
II Explain why in practice the reaction is carried out at 700K but NOT at 500K
(b) Hydrogen peroxide decomposes according to the following equation
: 2H₂O₂(aq) →2H₂O(l) + O₂ (g)
In an experiment, the rate of decomposition of hydrogen peroxide was found to be 6.0 x 10^-8 mol dm^-3 S^-1.
(i) Calculate the number of moles per dm³ of hydrogen peroxide that had decomposed within the first 2 minutes
(ii) In another experiment, the rate of decomposition was found to be 1.8 x 10^-7 mol dm^-3S^-1. The difference in two rates could have been caused by addition of a catalyst. State, giving reasons, one other factor that may have caused the difference in two rates of decomposition

ANSWERS