P1 kinetics help

[allthebest]

1.      Curve X on the graph below shows the volume of oxygen formed during the catalytic decomposition of a 1.0 mol dm–3 solution of hydrogen peroxide.

2H2O2(aq) → O2(g) + 2H2O(l)

 

           Which change would produce the curve Y? (picture attached)

A.        Adding water

B.        Adding some 0.1 mol dm–3 hydrogen peroxide solution

C.        Using a different catalyst

D.        Lowering the temperature

Why is the answer B?

I thought it was A since if we put more water, concentration of hydrogen peroxide will decrease? and if we add more hydrogen peroxide, shouldn't the rate increase since there are more of the reactant?

2. Powdered manganese(IV) oxide, MnO2(s), increases the rate of the decomposition reaction of hydrogen peroxide, H2O2(aq). Which statements about MnO2 are correct?

I.    The rate is independent of the particle size of MnO2.

II.   MnO2 provides an alternative reaction pathway for the decomposition with a lower activation energy.

III.  All the MnO2 is present after the decomposition of the hydrogen peroxide is complete.

A.        I and II only

B.        I and III only

C.        II and III only

D.        I, II and III

Does the particle size of the catalyst affect rate of reaction?

Thank You!

[IB`NOT`ez]

Not too sure about the second question as catalysts do appear in the rate expressions for certain reactions and as changing surface area also increases/decreases its rate of reaction, I should think the surface area of catalysts, provided they appear in the rate expression, would affect the rate of reaction. 

(Someone please correct me if I'm wrong) As for the first question, increasing the amount of reactants will shift the "plateau" of the curve upwards as more of the products can be formed from the increased amount of reactants. However, increasing the amount of reactants does not necessarily increase the rate of reaction, especially if we do not know what the order of reaction with respects to it is. Curve Y shows the upwards shift in the plateau, albeit with a slower rate of reaction; hence, H2O2 is likely to be zero order with respects to it.

From an enzymatic perspective in Biology: The saturation point, or plateau, is only attained when all the active sites of the limited amount of enzymes are in constant use. However by increasing the amount of enzymes, the saturation point can be shifted upwards as a result of more active sites being available for use. 

Edit: why did i waste time writing this when there were still no answers to the OP, only for the question to be answered 2 hours later and OP got her help whilst this post is still hidden 4 hours later

Edited August 3, 2016 by IB`ez

[kw0573]

1. The key recognition is that in the completion of reaction Y, more oxygen is produced. The ONLY way that happens is if you add more reactants. It does not appear that #1. is a rate question. 

2. The catalyst changes the k, rate constant. I didn't know this was in syllabus, but basically solid particles affect rate not by concentration but by surface area. If more solid increases the rate, then smaller particles mean higher rate, higher surface area, while larger particles mean lower rate. Now a similar effect take place with the solid catalyst. However, catalyst changes the k, rate constant, instead of appearing as a multiplier, but smaller particles mean higher rate regardless. I am not sure if you had done such an experiment at school, of stirring the reaction chamber, and thus decreasing the size of solid particles, to speed up the reaction.

[allthebest]

7 minutes ago, kw0573 said:

1. The key recognition is that in the completion of reaction Y, more oxygen is produced. The ONLY way that happens is if you add more reactants. It does not appear that #1. is a rate question. 

2. The catalyst changes the k, rate constant. I didn't know this was in syllabus, but basically solid particles affect rate not by concentration but by surface area. If more solid increases the rate, then smaller particles mean higher rate, higher surface area, while larger particles mean lower rate. Now a similar effect take place with the solid catalyst. However, catalyst changes the k, rate constant, instead of appearing as a multiplier, but smaller particles mean higher rate regardless. I am not sure if you had done such an experiment at school, of stirring the reaction chamber, and thus decreasing the size of solid particles, to speed up the reaction.

then for no.2 do you mean that the answer is wrong? you're saying particle size of the catalyst will increase the rate of reaction?

[kw0573]

3 minutes ago, allthebest said:

then for no.2 do you mean that the answer is wrong? you're saying particle size of the catalyst will increase the rate of reaction?

Of course not. All I say is that rate is NOT independent on size of MnO₂, which is what you want me to explain .
And like I said, particle size will be inversely proportional to rate, so there is dependency, so statement I out of the III is false. 

Edited August 3, 2016 by kw0573

[allthebest]

56 minutes ago, kw0573 said:

Of course not. All I say is that rate is NOT independent on size of MnO₂, which is what you want me to explain .
And like I said, particle size will be inversely proportional to rate, so there is dependency, so statement I out of the III is false. 

Ohhh ok ok thank you!!