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 :)
3 answers
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.
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!
1 answer