For the reaction below, describe the temperature and pressure conditions that would produce the highest yield for the forward reaction. Explain your answer in terms of Le Châtelier’s principle.

CO (g) + 2 H2 (g) Two arrows stacked on top of each other. The top arrow points to the right. The bottom arrow points to the left. CH3OH (g) deltaH = -90.7 kJ/mol

Question

For the reaction below, describe the temperature and pressure conditions that would produce the highest yield for the forward reaction. Explain your answer in terms of Le Châtelier’s principle. 

CO (g) + 2 H2 (g) Two arrows stacked on top of each other. The top arrow points to the right. The bottom arrow points to the left. CH3OH (g) deltaH = -90.7 kJ/mol

anonymous · Answer

@Frostbite think you could help me out here?

anonymous · Answer

@Luigi0210 @Compassionate think you could help?

anonymous · Answer

@Loser66

anonymous · Answer

@bookworm00981

frostbite · Answer

Okay it is a reversible reaction, what $\rightleftharpoons$ means.
Now Le Châtelier’s principle states that if you try to affect a equilibrium, it will cause a change that try lower the effect of the action. (sorry for the very poor explanation)
What this means for your reaction will be explained below:

frostbite · Answer

Lets look at the following reaction:
$$\Large \sf CO~(g) + 2 ~ H _{2} (g) \rightleftharpoons CH _{3}OH (g)$$
The equilibriumconstant for this reaction is:
$$\Large K=\frac{ \sf \left[ CH _{3}OH ~ (g) \right] }{ \left[ \sf CO ~ (g) \right] \times \left[ \sf H _{2} (g) \right]^{2} }$$
Lets increase the pressure, which we do by lowing the volume, the gasconcentration then increase. Avogadro's law still apply for the system so all the concentrations is raised $\huge \sf equally $ much. It then follows from the equilibrium constant the system is no longer in equilibrium. The denominator is almost 2x as high as the numerator. The way the system will make for this is to go towards the right.

anonymous · Answer

OK so when it asks for all that stuff like " temperature and pressure conditions that would produce the highest yield for the forward reaction" what should i say?

frostbite · Answer

If that reaction if you want to get the highest yield you increase the pressure.
For temperature we want to lower it as the enthalpy for the reaction is negative.

anonymous · Answer

ok thank you and one last question 

Balance the following redox equation and identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent. Show all of the work used to solve the problem. 

CN- + MnO4- yields CNO- + MnO2

"Cu + 2 HNO3 + 6H+ >> 3Cu2+ + 2NO + 4 H2O 

CN- + H2O >> CNO- + 2H+ + 2e-) x3 

MnO4-+ 4H+ + 3e- >> MnO2 + 2 H2O) x2 " is that correct?