# Chemistry

posted by .

Given the equilibrium: CO + 2H2 = CH3OH
H=-18.0kJ
How will the concentration of CO at equilibrium be affected by the following?

b) Removing some H2
c) Reducing the temperature.

• Chemistry -

The long explanation is that when a system at equilibrium is put under stress (T, P, C, catalyst, etc), the system will shift in such as way so as to relieve the stress. Here is the simple explanation.
A system at equilibrium will try to undo what we do to it.
So let me rewrite the equation and I will ASSUME that all of the reactants and products are gases.
CO + 2H2 = CH3OH + heat.
Note that I have added the heat term into the equation instead of two stand alone items. Another note. There are only three answers to problems like this.
a. The reaction shifts to the left.
b. The reaction shifts to the right.'
c. No effect (the reaction doesn't shift either way). The only time you see c is when the number of moles of gas are the same on both sides of the equation.
So adding CH3OH, the system will try to undo that. What must it do to undo what I did by adding CH3OH? It will try to take away CH3OH. How can it do that? The equilibrium can shift to the LEFT because that uses up CH3OH and forms more CO and more H2. (Of the three answers, you KNOW it can't shift to the right because that forms MORE CH3OH. You KNOW it can't stay the same because the moles are not the same on both side. The only answer that makes sense is a), shift to the left because that uses up the added CH3OH.
b)Removing some H2. So we take away some H2. The system will try to undo that. So it will try to ADD H2. The ONLY way it can do that is to shift to the left. Shifting to the right would use up H2.
I'll leave c for you. Reducing T means taking away heat or making it colder. So the reaction will ......

## Similar Questions

1. ### Chemistry

Methanol (CH3OH) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: CO (g) + 2H2 (g) -> CH3OH (g) An equilibrium mixture in a 2.50 L vessel is found to contain 2.62e-2 mol CH3OH, 0.170 mol CO, and …
2. ### chemistry

1.2 mol of CH3OH(g) are injected into a 2.0 L container and the following equilibrium becomes established. 2H2(g) + CO(g) <=====> CH3OH(g) + 92 kJ If at equilibrium 1.0 mol of CH3OH is still in the container the Ke must be which …
3. ### chemistry

at 327 degrees C, the equilibrium concentrations are [CH3OH]=0.15M, [CO]=0.24M, and [H2]=1.1M for the reaction: CH3OH(g)<-->CO(g)+2H2(g) Calculate Kp at this temperature
4. ### College Chemistry 2 - Equilibrium Constants/Conc.

The equilibrium constant for the reaction is 37 at a certain temperature. 2H2(g) + CO(g)<==> CH3OH(g) If there are .0293 moles of H2 and .00353 moles of CH3OH at equilibrium in a 4.53-L flask, what is the concentration of CO?
5. ### chem

At a certain temperature, the reaction CO(g) + 2H2(g) <===> CH3OH(g) has Kc = 0.500. If a reaction mixture at equilibrium contains 0.00436 M CH3OH and 0.220 M H2, what is the equilibrium concentration of CO?
6. ### chemistry

equilibrium CO(g)+2H2(g)<->CH3OH(g)calculate these concentrations:CO(g) in an equilibrium mixture containing 0.933mol/L H2 and 1.32 mol?
7. ### chemistry

For the equilibrium system below, which of the following would result in an increase in the concentration of CO(g)?
8. ### chemistry

the reaction CO + 2H2<----->CH3OH has a delta H -18 kJ. how will the amount of CH3OH be affected by the following. a) removing H2(g) b) decreasing the volume of the container.
9. ### Chemistry1412

Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) An equilibrium mixture of this reaction at a certain temperature was found to have [CO]= 0.115M , [H2]= 0.110M , and [CH3OH]= 0.190M . What is the value of the equilibrium …
10. ### Chemistry

Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a specific temperature with initial concentrations of [CO] = 0.27 M and [H2] = 0.49 M. At equilibrium, the concentration of CH3OH is 0.11 …

More Similar Questions