Answer:
To increase the yield of H₂ we would use a low temperature.
For an exothermic reaction such as this, decreasing temperature increases the value of K and the amount of products at equilibrium. Low temperature increases the value of K and the amount of products at equilibrium.
Explanation:
Let´s consider the following reaction:
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)
When a system at equilibrium is disturbed, the response of the system is explained by Le Chatelier's Principle: <em>If a system at equilibrium suffers a perturbation (in temperature, pressure, concentration), the system will shift its equilibrium position to counteract such perturbation</em>.
In this case, we have an exothermic reaction (ΔH° < 0). We can imagine heat as one of the products. If we decrease the temperature, the system will try to raise it favoring the forward reaction to release heat and, at the same time, increasing the yield of H₂. By having more products, the value of the equilibrium constant K increases.
Boiling point elevation is given as:
ΔTb=iKbm
Where,
ΔTb=elevation in the boiling point
that is given by expression:
ΔTb=Tb (solution) - Tb (pure solvent)
Here Tb (pure solvent)=118.1 °C
i for CaCO3= 2
Kb=2.93 °C/m
m=Molality of CaCO₃:
Molality of CaCO₃=Number of moles of CaCO₃/ Mass of solvent (Kg)
=(Given Mass of CaCO3/Molar mass of CaCO₃)/ Mass of solvent (Kg)
=(100.0÷100 g/mol)/0.4
= 2.5 m
So now putting value of m, i and Kb in the boiling point elevation equation we get:
ΔTb=iKbm
=2×2.93×2.5
=14.65 °C
boiling point of a solution can be calculated:
ΔTb=Tb (solution) - Tb (pure solvent)
14.65=Tb (solution)-118.1
Tb (solution)=118.1+14.65
=132.75