Help with P1 Kinetics

[allthebest]

1. At 25 °C, 200 cm3 of 1.0 mol dm–3 nitric acid is added to 5.0 g of magnesium powder. If the experiment is repeated using the same mass of magnesium powder, which conditions will result in the same initial reaction rate?

 

	Volume of HNO3 / cm3	Concentration of HNO3 / mol dm–3	Temperature / °C
A.	200	2.0	25
B.	200	1.0	50
C.	100	2.0	25
D.	100	1.0	25

ANS: D

= I don't get when there are more than 1 variables... Can anyone explain which one is dominant or what happens when both volume and concentration is like changed like this question?

2.  Which experimental procedure could be used to determine the rate of reaction for the reaction between a solution of cobalt chloride, CoCl2(aq), and concentrated hydrochloric acid, HCl(aq)?

 

                         Co(H2O)62+(aq) + 4Cl–(aq)  CoCl42–(aq) + 6H2O(l)

 

A.      Measure the change in pH in a given time

B.      Measure the change in mass in a given time

C.      Use a colorimeter to measure the change in colour in a given time

D.      Measure the change in volume of the solution in a given time

ANS: C

= how do you know that we need to use colorimeter?

3. 

Sodium thiosulfate solution, Na2S2O3(aq), and hydrochloric acid, HCl(aq), react spontaneously to produce solid sulfur, S(s), according to the equation below.

                                    S2O32–(aq) + 2H+(aq) → S(s) + SO2(aq) + H2O(l)

          A student experimentally determined the rate expression to be:

                                                                 rate = k[S2O32–(aq)]2

          Which graph is consistent with this information?

* pic attached

ANS: B

I thought is was D... Isn't this second order reaction?

4. 

Equal masses of powdered calcium carbonate were added to separate solutions of hydrochloric acid. The calcium carbonate was in excess. The volume of carbon dioxide produced was measured at regular intervals. Which curves best represent the evolution of carbon dioxide against time for the acid solutions shown in the table below.

         

 *Pic attached

 

	25 cm3 of 2 mol dm–3 HCl	50 cm3 of 1 mol dm–3 HCl	25 cm3 of 1 mol dm–3 HCl
A.	I	III	IV
B.	I	IV	III
C.	I	II	III
D.	II	I	III

 

ANS: C

I think this one is similar to the first one. When does the final volume change? When does the final volume halve?

[kw0573]

1. Volume does not impact rate for liquids and solids and aqueous solutions because matter in these states matter is not visibly compressible. So same temperature, same concentration would be D. However for gases you need to consider if pressure or temperature have changed due to change in volume, and then we need to consider changes of pressure and temperature on rate. 

2. You have to know about transition metals and complex ions, which is probably a later unit for you. The metals' d-orbitals are split, and when electrons from the high energy d-orbitals de-excite to the ground state, it releases electromagnetic waves, which happen to be mostly in the visible region. This depend on what the transition metal is bonded to, such as water or chloride.

3. Be careful what are the x- and y- axes. If rate = k * [concentration]^2, then a rate-concentration graph should be quadratic. It is akin to y = kx^2. Choice D shows first order, or rate = k * [concentration]. 

4. This is both a stoichiometry and rate question, so not quite the same as the pure rate question in #1. This is a double displacement reaction (producing H₂CO₃, which spontaneously produces H₂O and CO₂). If on the test you know the reaction, you should write it down.
CaCO₃ + 2HCl --> CaCl₂ + H₂O + CO₂. 
The big idea is that more HCl as starting material will produce more CO₂ in the end, and this observation hopefully is evident even without writing down the reaction. 

mol = concentration * volume. Since we are not asked for actual number of moles, you can just directly multiply concentration and volume, without worrying about compatible units.
 

25 cm3 of 2 mol dm–3 HCl => 50 mol dm-3 cm3

50 cm3 of 1 mol dm–3 HCl => 50 mol dm-3 cm3

25 cm3 of 1 mol dm–3 HCl => 25 mol dm-3 cm3

So chambers 1 and 2 produce twice as much produce as chamber 3. This already narrows the answer to C and D. 
Now comparing rates, we see chambers 2 and 3 have same initial concentration, which is less than that of chamber 1, so chamber I will have a higher instantaneous rate than chambers 2 and 3.

Looking at C and D, C is the correct if we just look at chambers 1 and 2. [If you are in a time constraint, it would generally be sufficient to stop here on the actual exam and choose C.] But is graph III correct for chamber 3?
Rate law is for instantaneous change in volume with respect to change in time. Mathematically this is called a derivative and is the slope of the tangent to the volume-concentration graph. We only know (without further calculations) about the rate at t = 0, because we know the concentration of the reactant at the moment. You can see that II and III have same tangent at t = 0. Because chamber 3 has half the amount (mole) of HCl compared to chamber 2, that it should make sense that chamber 3 reaches completion before chamber 2. 

 

 

[Msj Chem]

 

Edited June 19, 2016 by Msj Chem

[allthebest]

13 hours ago, kw0573 said:

1. Volume does not impact rate for liquids and solids and aqueous solutions because matter in these states matter is not visibly compressible. So same temperature, same concentration would be D. However for gases you need to consider if pressure or temperature have changed due to change in volume, and then we need to consider changes of pressure and temperature on rate. 

2. You have to know about transition metals and complex ions, which is probably a later unit for you. The metals' d-orbitals are split, and when electrons from the high energy d-orbitals de-excite to the ground state, it releases electromagnetic waves, which happen to be mostly in the visible region. This depend on what the transition metal is bonded to, such as water or chloride.

3. Be careful what are the x- and y- axes. If rate = k * [concentration]^2, then a rate-concentration graph should be quadratic. It is akin to y = kx^2. Choice D shows first order, or rate = k * [concentration]. 

4. This is both a stoichiometry and rate question, so not quite the same as the pure rate question in #1. This is a double displacement reaction (producing H₂CO₃, which spontaneously produces H₂O and CO₂). If on the test you know the reaction, you should write it down.
CaCO₃ + 2HCl --> CaCl₂ + H₂O + CO₂. 
The big idea is that more HCl as starting material will produce more CO₂ in the end, and this observation hopefully is evident even without writing down the reaction. 

mol = concentration * volume. Since we are not asked for actual number of moles, you can just directly multiply concentration and volume, without worrying about compatible units.
 

25 cm3 of 2 mol dm–3 HCl => 50 mol dm-3 cm3

50 cm3 of 1 mol dm–3 HCl => 50 mol dm-3 cm3

25 cm3 of 1 mol dm–3 HCl => 25 mol dm-3 cm3

So chambers 1 and 2 produce twice as much produce as chamber 3. This already narrows the answer to C and D. 
Now comparing rates, we see chambers 2 and 3 have same initial concentration, which is less than that of chamber 1, so chamber I will have a higher instantaneous rate than chambers 2 and 3.

Looking at C and D, C is the correct if we just look at chambers 1 and 2. [If you are in a time constraint, it would generally be sufficient to stop here on the actual exam and choose C.] But is graph III correct for chamber 3?
Rate law is for instantaneous change in volume with respect to change in time. Mathematically this is called a derivative and is the slope of the tangent to the volume-concentration graph. We only know (without further calculations) about the rate at t = 0, because we know the concentration of the reactant at the moment. You can see that II and III have same tangent at t = 0. Because chamber 3 has half the amount (mole) of HCl compared to chamber 2, that it should make sense that chamber 3 reaches completion before chamber 2. 

 

 

Thank you very much for clarifying why I got those answers wrong. It really helped me.