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Komok [63]
4 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]4 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]4 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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A 0.75M solution of CH3OH is prepared in 0.500 kg of water. How many moles of CH3OH are needed?
4vir4ik [10]

Answer:

We need 0.375 mol of CH3OH to prepare the solution

Explanation:

For the problem they give us the following data:

Solution concentration 0,75 M

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knowing that the density of water is 1g / mL,  we find the volume of water:

                           d = \frac{g}{mL} \\\\ V= \frac{g}{d}  = \frac{500g}{1 \frac{g}{mL} } = 500mL = 0,5 L

Now, find moles of CH_{3} OH are needed using the molarity equation:

                           M = \frac{ moles }{ V (L)} \\\\\\molesCH_{3}OH  = M . V(L) = 0,75 M . 0,5 L\\\\molesCH_{3}OH = 0,375 mol

therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH,  resulting in a concentration of 0,75M

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