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Komok [63]
3 years ago
7

Match the effects on the concentration of SO3 gas when the following changes occur after initial equilibrium has been establishe

d in this system (N.C. = no change). 2SO2(g) + O2(g) 2SO3(g) + 46.8 kcal pressure increases increase [O2] increase volume of chamber adding a catalyst raising the temperature
Chemistry
2 answers:
Archy [21]3 years ago
8 0

The reaction is

2SO_{2} (g)+O_{2}(g)->2SO_{3} (g)+46.8 kcal

1) the system is homogeneous with all the reactants and products are gases

2) there are three moles on reactant side and two moles on product side

i) If we increase pressure then there will be decrease in volume. It will cause increase in moles per unit volume. So if we increase pressure the equilibrium will shift towards side having less number of moles of gaseous species.

Thus system will move in forward direction.

ii) If we increase concentration of oxygen it means we are increasing concentration of reactants,this will cause system to shift in forward direction

iii) If we increase volume of chamber,  it will cause decrease in moles per unit volume, hence the system will move in the direction where number of moles of gaseous species is more, System will shift in reverse direction

iv) There is generally no effect of catalyst on equilibrium. However an increase in temperature will cause shift in reverse direction.

mestny [16]3 years ago
3 0

The effects on the concentration of SO3 gas when the following changes occur after initial equilibrium has been established in this system (N.C. = no change) by adding a catalyst.

2SO2(g) + O2(g)  2SO3(g) + 46.8 kcal

So that even though there is an addition of catalyst, no change in reactants or products has occurred because catalyst only provides a faster pathway for the reaction to occur.

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4 0
2 years ago
What is the theoretical yield of methanol (CH3OH) when 12.0 grams of H2 is mixed with 74.5 grams of CO? CO + 2H2 CH3OH
AlladinOne [14]
1) We need to convert 12.0 g of H2 into moles of H2, and <span> 74.5 grams of CO into moles of CO
</span><span>Molar mass of H2:    M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO:   M(CO) = 12.0 +16.0 = 28.0 g/mol

</span>12.0 g  H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO

<span>2) Now we can use reaction to find out what substance will react completely, and what will be leftover. 

                                  CO       +         2H2   ------->      CH3OH  
                                 1 mol              2 mol
given                        2.66 mol          6 mol (excess)

How much
we need  CO?           3 mol              6 mol

We see that H2 will be leftover, because for 6 moles H2  we need 3 moles CO, but we have only 2.66 mol  CO.
So, CO will react completely, and we are going to use CO to find  the mass of CH3OH.

3)                              </span>CO       +         2H2   ------->      CH3OH  
                                 1 mol                                        1 mol
                                2.66 mol                                    2.66 mol

4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 +  4*1.0 + 16.0 = 32.0 g/mol

2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH =  85.12 g CH3OH
<span>
Answer is </span>D) 85.12 grams.
3 0
2 years ago
Please help me. Quickly!!!
Tomtit [17]

Answer:

500

Explanation:

8 0
2 years ago
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