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.
The answer for the following problem is mentioned below.
- <u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>
Explanation:
Given:
no of moles of the oxygen gas = 0.692
Also given:
2 HgO → 2 Hg + 
where,
HgO represents mercuric oxide
Hg represents mercury
represents oxygen
To calculate:
Molar mass of HgO:
Molar mass of HgO = 216 grams
molar mass of mercury (Hg) = 200 grams
molar mass of oxygen (O) =16 grams
HgO = 200 +16 = 216 grams
We know;
2×216 grams of HgO → 1 mole of oxygen molecule
? → 0.692 moles of oxygen molecule
= 
= 298.944 grams of HgO
<u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>
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The answer here is letter C. The optimal solution. The optimal solution is the one that affects how certain things changes with sensitivity analysis. The optimal solution is a feasibility solution where the objective function of it is to reach the minimum and maximum value.
