questions in the picture below (show your works) For the following reaction to m
ID: 1005429 • Letter: Q
Question
questions in the picture below (show your works)
For the following reaction to make glucose: 6 CO_2 (g) + 6 H_20 (1) doubleheadarrow C_6H_12O_6, (s) + 6 O_2 (g) deltaH degree = + 2,816 kJ. How is the equilibrium yield of glucose affected by each ot the following. Increasing partial pressure of carbon dioxide ? Increasing temperature ? Removing carbon dioxide ? Decreasing the total pressure ? Removing part of the glucose ? Adding a catalyst ? Decreasing the total pressure with an inert gas ? Briefly explain your answer in part (ii) Briefly explain your answer in part (vi) Briefly explain your answer in part (vii)Explanation / Answer
The equilibrium constant
K= [O2]6/ [CO2]6.
The reaction is endothermic, that is heat is absorbed during the course of the reaction. So when temperature is increased, the reaction as per Lechatlier principle proceeds in a direction so as to compensate the effect of temperature, i.e endothermic direction. So more of glucose is formed.
When CO2 partial pressure is increased, denominator in the K terms increase and O2 pressure also increases. So the reaction proceeds in a direction of decreasing moles. So more of Glucose is formed.
When CO2 is removed, there is a decrease in moles on the reactant side. So the reaction shits to the reactants side and less C6H12O6.
Decreasing the total pressure will not have any effect since there are 6 moles of gas on the reactant side and 6 moles of gas on the product side .
When glucose is removed, there is a decrease in moles of product side and hence the reaction proceeds in a direction where there is an increase in moles. The reaction proceeds to the left.
Inert gas does not form a part of equilibrium constant and hence does not have any affect.
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