# Chemistry question help

Hi, Could use some help with the last part of this question.
The curve below shows how the volume of oxygen evolved varies with time when 50
cm' of a 2.0 mol dm solution of hydrogen peroxide, 402, decomposes at 298 K.
(a)
(b)
Volume
of oxygen
/ cm3
Time f s
State how you could use the curve to find the rate of reaction at point A.
(1)
Sketch curves, on the above axes, to illustrate how the volume of oxygen evolved
would change with time if the experiment was repeated at 298 K using the following.
100 cm! of a 1.0 mol dm solution of HO. Label this curve X.
25 cm! of a 2.0 mol dm solution of H202 in the presence of a catalyst.
Label this curve Y.
(4)

(b)
Curve X is lower and starts at origin
And levels out at same volume as original curve
Curve Y is steeper than original and starts at origin
Then levels out at half the volume of the original

Why is curve Y steeper if they have the same concentrations?
That repeat is done with the additional presence of a catalyst. Catalysts speed up the rate of reaction, so the rate of decomposition of H2O2 would increase. Hence there would be a greater rate of oxygen production (larger volume of oxygen produced per unit time), which is reflected by the graph being steeper.
(edited 1 month ago)
Original post by Methene
That repeat is done with the additional presence of a catalyst. Catalysts speed up the rate of reaction, so the rate of decomposition of H2O2 would increase. Hence there would be a greater rate of oxygen production (larger volume of oxygen produced per unit time), which is reflected by the graph being steeper.

Thanks! That makes sense! Does this mean that without a catalyst the curves should be identical because concentrations are identical?
Original post by mitostudent
Thanks! That makes sense! Does this mean that without a catalyst the curves should be identical because concentrations are identical?

Without the catalyst, it would have the same steepness but still levels out at half the volume of the original. One other thing to consider when dealing with rate curves is total moles of reactant. The original curve is for 50cm3 of 2M solution, which is 0.01 moles. Curve Y is for 25cm3 of 2M solution, which is 0.005 moles. That's why it levels out at half the volume, as it can only make half the product compared to original.

You can check this reasoning with curve X
Original post by Methene
Without the catalyst, it would have the same steepness but still levels out at half the volume of the original. One other thing to consider when dealing with rate curves is total moles of reactant. The original curve is for 50cm3 of 2M solution, which is 0.01 moles. Curve Y is for 25cm3 of 2M solution, which is 0.005 moles. That's why it levels out at half the volume, as it can only make half the product compared to original.
You can check this reasoning with curve X

awesome! thanks