osteoblogda

2022-01-03

The water-gas shift reaction
$CO\left(g\right)+{H}_{2}O\left(g\right)⇋C{O}_{2}\left(g\right)+{H}_{2}\left(g\right)$
is used industrially to produce hydrogen. The reaction enthalpy is
$\mathrm{\Delta }{H}^{\circ }=-41kJ$
To increase the equilibrium yield of hydrogen would you use high or low temperature?

lovagwb

Step 1
Given water-gas shift reaction:
$CO\left(g\right)+{H}_{2}O\left(g\right)⇋C{O}_{2}\left(g\right)+{H}_{2}\left(g\right)$
$\mathrm{\Delta }H$ for this reaction is negative, that means reaction is exothermic and reversible.
Now, as per Le Chatlier's principle, for exothermic reaction, when the temperature is increased then the equilibrium shifts towards the left
Therefore, products will get less formed and the reaction is thermodynamically favored at low temperatures and kinetically favored at high temperatures.
Hence, at low temperature, the equilibrium yield of hydrogen gas will be more.
At low temperature, the equilibrium yield of hydrogen gas will be more.

Steve Hirano

Step 1
For the reaction:
$CO\left(g\right)+{H}_{2}O\left(g\right)⇋C{O}_{2}\left(g\right)+{H}_{2}\left(g\right)$
$\mathrm{\Delta }{H}^{\circ }=-41kJ$
As the reaction is exothermic $\left(\mathrm{\Delta }{H}^{\circ }<0\right)$, you need to use low temperature to increase the equilibrium yield of hydrogen -LeChatelier's principle-.
We would use low temperature. For an exothermic reaction such as this, decreasing temperature increases the value of K and the amount of products at equilibrium.

karton

Step 1
Explanation:
Let's consider the following reaction:
$CO\left(g\right)+{H}_{2}O\left(g\right)⇌C{O}_{2}\left(g\right)+{H}_{2}\left(g\right)$
When a system at equilibrium is disturbed, the response of the system is explained by Le Chatelier's Principle: If a system at equilibrium suffers a perturbation (in temperature, pressure, concentration), the system will shift its equilibrium position to counteract such perturbation.
In this case, we have an exothermic reaction $\left(\mathrm{\Delta }{H}^{\circ }<0\right)$. 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.

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