## Photon336 one year ago Question

1. Photon336

Questions below

2. Photon336

@Rushwr

3. taramgrant0543664

Wouldn't it be A? Products over reactants, stoiciometric coefficients become powers

4. taramgrant0543664

For the second one if you removes CH4 then the reaction would want to go to the left. Adding CO would drive the reaction left as well. Removing H2 would cause the reaction to be driven right causing H2 to be produced

5. taramgrant0543664

As for pressure I couldn't remember what principle it was but its Le Chatelier's Principle and pressure will Try to even out the number of moles the right side has 4 and the left side has 2 so increasing the pressure would drive the reaction left so B is the answer for the second one

6. Rushwr

For the 1st question it is A For the 2nd question it is B. A is wrong cuz if we add CO the system will produce more CH4 and H2O in order to regain its equilibrium. B is correct cuz when we remove H2 the system is lacking H2 and it effects te equilibrium hence it tends to form more H2. C is wrong , if we remove CH4 that is we are removing a reactant . In order to regain its eqm. CH4 is formed favoring the reverse reaction. D is wrong . U see that the total moles in the product side is 4 while the total moles in the reactant side is 2 hence if we increase the pressure the equilibrium will shift to left side again.

7. Photon336

$NO _{2}(g) + O _{2}(g) \rightarrow 2NO _{2}(g)$

8. Photon336

$\frac{ [NO2]^{2} }{ [NO]^{2} [O2] } = K _{eq}$

9. Photon336

We only consider gases in the equilibrium expression, rather than solids and pure liquids because the concentration of solids, & pure liquids doesn't change much.

10. Photon336

the first one is A

11. Photon336

2 is B

12. Photon336

removing $H _{2}$ from the reaction will cause it to go to the right.

13. Photon336

will post another!