Methanol is manufactured from carbon (IV) oxide and hydrogen gas according to the equation below: The reaction is carried out in the presence of a chromium catalyst...

      

Methanol is manufactured from carbon (IV) oxide and hydrogen gas according to the
equation below:
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The reaction is carried out in the presence of a chromium catalyst at 700K and 30kPa. Under
these conditions, an equilibrium is reached when 2 % of the carbon (iv) oxide is converted to
methanol?
(a) How does the rate of forward reaction compare with that of the reverse reaction
when 2% of the carbon(iv)oxide is converted to methanol?
(b) Explain how each of the following conditions would affect the yield of methanol:
(i) Reduction in pressure
(ii) Using a more efficient catalyst.
(c) If the reaction is carried out at 500K and 30kPa the percentage of carbon (iv) oxide
converted is higher than 2%. What is the sign of ΔH for the reaction? Explain.

  

Answers


sharon
(a) The rate of the forward reaction is higher than that of backward reaction.
(b) (i) More CO2 and H2 would be formed.
This involved in increased in volume /number of molecules from 2 to 4
molecules.
(ii) Methanol would be produced faster;since the catalyst would establish
the equilibrium faster.
(c) Negative-Reduction in temperature favours an exothermic reaction.
sharon kalunda answered the question on May 22, 2019 at 07:26


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