CO(g) + 2H_2(g) <==> CH_3OH (g) deltaH=-90.7kJ Methanol is prepared industrially from synthesis gas (CO and H_2)
Would the fraction of methanol obtained at equilibrium be increased by raising the temperature? Explain.
Thank you :)
Oh I made a mistake. Instead of -90.7kJ in delta H, it should've been -21.7kcal. Thank you so much
This is a problem involving Le Chatelier's Principle. dH is negative which means the process is exothermic. I like to rewrite the equation this way.
CO(g) + 2H2(g) <==> CH3OH(g) + heat
Le Chatelier's Principle says that a system at equilibrium, when subjected to a stress, will shift (to the right or left) to undo what we did to it. Therefore, if the system is heated, it will try NOT to produce more heat so less CH3OH will be produced.
thank you so much!
To determine whether raising the temperature would increase the fraction of methanol obtained at equilibrium, we need to consider the effect of temperature on the equilibrium position.
According to Le Chatelier's principle, when a system at equilibrium is subjected to a change, the system will adjust to counteract the change and maintain equilibrium. In this case, increasing the temperature is considered a change in the system.
Let's take a closer look at the given reaction: CO(g) + 2H2(g) <==> CH3OH(g)
The synthesis of methanol is an exothermic reaction, as indicated by the negative deltaH value (deltaH = -90.7 kJ/mol). This means that the forward reaction, converting CO and H2 into methanol, releases heat.
According to Le Chatelier's principle, when the temperature is increased, the system will shift in the direction that consumes heat. In this case, the forward reaction is exothermic. Therefore, increasing the temperature would shift the equilibrium to the left, favoring the reactants (CO and H2) over the products (CH3OH).
Consequently, raising the temperature would decrease the fraction of methanol obtained at equilibrium because the equilibrium position would shift towards the reactants. Therefore, the answer to the question is no, the fraction of methanol obtained at equilibrium would not be increased by raising the temperature.
I hope this explanation helps! Let me know if you have any further questions.