yield vs. time

[dinosaur93]

Say I have an equation like this:

A + B ----> C


When I increase the concentration of molecule A, the number of successful collision would increase thereby increasing the rate of reaction yeah? therefore more products are being formed.....BUT when is it that the yield would stay the same even though I have added extra reactants?

[thushan]

Say I have an equation like this:

A + B ----> C


When I increase the concentration of molecule A, the number of successful collision would increase thereby increasing the rate of reaction yeah? therefore more products are being formed.....BUT when is it that the yield would stay the same even though I have added extra reactants?

If the reaction were complete.

[dinosaur93]

Say I have an equation like this:

A + B ----> C


When I increase the concentration of molecule A, the number of successful collision would increase thereby increasing the rate of reaction yeah? therefore more products are being formed.....BUT when is it that the yield would stay the same even though I have added extra reactants?

If the reaction were complete.

How about if the reaction does not go completely react...

say for instance, If I only increase the concentration of one reactant, would that still be sufficient to increase the yield of the overall reaction, or does it stays the same?

[thushan]

The answer to that question is REALLY complicated.

Say A + B <--> C, let's take two cases:

CASE 1: A originally in excess.

If this is the case, B is the limiting reagent and therefore the yield will be the amount of B converted to C.
If we add extra A, more B would be converted to C, and the yield will increase.
If we add extra B (keeping A in excess), yes the position of equilibrium will shift forward, but the percentage of B converted to C will be less, so yield actually decreases.

CASE 2: B originally in excess.

If this is the case, A is the limiting reagent and therefore the yield will be the amount of A converted to C.
If we add extra B, more A would be converted to C, and the yield will increase.
If we add extra A (keeping B in excess), yes the position of equilibrium will shift forward, but the percentage of A converted to C will be less, so yield actually decreases.

That's why the answer to that is very complicated.

[dinosaur93]

The answer to that question is REALLY complicated.

Say A + B <--> C, let's take two cases:

CASE 1: A originally in excess.

If this is the case, B is the limiting reagent and therefore the yield will be the amount of B converted to C.
If we add extra A, more B would be converted to C, and the yield will increase.
If we add extra B (keeping A in excess), yes the position of equilibrium will shift forward, but the percentage of B converted to C will be less, so yield actually decreases.

CASE 2: B originally in excess.

If this is the case, A is the limiting reagent and therefore the yield will be the amount of A converted to C.
If we add extra B, more A would be converted to C, and the yield will increase.
If we add extra A (keeping B in excess), yes the position of equilibrium will shift forward, but the percentage of A converted to C will be less, so yield actually decreases.

That's why the answer to that is very complicated.

err....I see...Im a bit lost with this part..

'If we add extra A (keeping B in excess), yes the position of equilibrium will shift forward, but the percentage of A converted to C will be less, so yield actually decreases.'

how would the percentage of A converted decrease?

-Do we need to know this? As in like....is it an examinable section in the course?

[thushan]

They could examine this, its unlikely though.

Suppose you had, at equilibrium:

20 units of A
40 units of B
800 units of C

(so K = 1)

The maximum amount of C that could have formed is 820 units (if ALL of A has reacted), so the yield is 80/82 = 97.6% (in fractional form).

Suppose we added another 20 units of A.

Initially it will be

40 units of A
40 units of B
800 units of C

Using CAS (just for arguments sake!), at equilibrium we get

28.5 units of A
28.5 units of B
811.5 units of C

Perceentage yield = 811.5/840 = 96.6%, less than last time.

[dinosaur93]

They could examine this, its unlikely though.

Suppose you had, at equilibrium:

20 units of A
40 units of B
800 units of C

(so K = 1)

The maximum amount of C that could have formed is 820 units (if ALL of A has reacted), so the yield is 80/82 = 97.6% (in fractional form).

Suppose we added another 20 units of A.

Initially it will be

40 units of A
40 units of B
800 units of C

Using CAS (just for arguments sake!), at equilibrium we get

28.5 units of A
28.5 units of B
811.5 units of C

Perceentage yield = 811.5/840 = 96.6%, less than last time.

ah, made so much sense, thank you!

[stephanieteddy]

So sorry to but in! How did you get this part?

Using CAS (just for arguments sake!), at equilibrium we get

28.5 units of A
28.5 units of B
811.5 units of C

Perceentage yield = 811.5/840 = 96.6%, less than last time.

How would I calculate that using my CAS? Thank you!!