Chemistry 3/4 2013 Thread

[Edward21]

1) With ethanol, can it undergo a substitution reaction with HBr to form Bromoethane and water? I thought that chloroalkanes could only reaction with the OH- ion to form the alcohol, not the other way around? Or according to this question.. 

2) With the successive ionisation of H3PO4 in water, why do the following weak acids ionise to a lesser extent following the first one?

[lzxnl]

1)
Yes it can. It's slightly different here; chloroalkanes are reacting with a hydroxide ion, while the reaction between ethanol and hydrogen bromide is forming water. The hydroxide ion is less stable than the chloride ion as can be seen by their base strengths, so it will latch onto the carbon more readily than the chloride ion. However, in the reaction between ethanol and HBr, the alcohol group is protonated by the acidic hydrogen in HBr to form a R-OH2 + group, in which the water can drop off favourably for the bromide ion to bond to the carbon. Hydroxide ions do not appear anywhere here.
The first reaction is conducted in basic solution. The second reaction is conduced in neutral or acidic solution. They are not direct opposites of each other.

2) When phosphoric acid is deprotonated, it bears a negative charge. To become deprotonated again, you're asking for a negative charge to lose another positive charge. That is inherently energetically unfavourable. 

[Edward21]

1)
Yes it can. It's slightly different here; chloroalkanes are reacting with a hydroxide ion, while the reaction between ethanol and hydrogen bromide is forming water. The hydroxide ion is less stable than the chloride ion as can be seen by their base strengths, so it will latch onto the carbon more readily than the chloride ion. However, in the reaction between ethanol and HBr, the alcohol group is protonated by the acidic hydrogen in HBr to form a R-OH2 + group, in which the water can drop off favourably for the bromide ion to bond to the carbon. Hydroxide ions do not appear anywhere here.
The first reaction is conducted in basic solution. The second reaction is conduced in neutral or acidic solution. They are not direct opposites of each other.

2) When phosphoric acid is deprotonated, it bears a negative charge. To become deprotonated again, you're asking for a negative charge to lose another positive charge. That is inherently energetically unfavourable.

Thank-you so much, could you tell I went back to organic chem? You are a Chemistry God! 

[lolipopper]

question:

what will with eqilibrium units be for this:

CH3COOH + H20 --> CH3COO- + H30+

i get the answer: none. because (M X M)/(M X M) = 1. The answer at back is M.

[jgoudie]

question:

what will with eqilibrium units be for this:

CH3COOH + H20 --> CH3COO- + H30+

i get the answer: none. because (M X M)/(M X M) = 1. The answer at back is M.

This is because the you are calculating Ka not K.  Ka is the equilibrium constant with water included. 

Ka = [CH3COO-][H3O]/[CHOOH] thus has the unit M

K = [CH3COO-][H3O+]/[H2O][CHOOH] thus would have no unit as you said.

Hope this makes sense.  I have just updated the chemsiode youtube channel with acid/base equilibria videos also.  See the link below.

http://www.youtube.com/mrjasongoudie.

[Limista]

2) the rates would be unchanged as the concentration of substances have not changed, volume stays the same.  The addition of an inert gas does not affect the reaction.

actually the rate of the forward reaction would decrease, because there is a decrease in the number of collisions per second between reactant molecules, thus slowing the rate. However, the concentrations of the reactant and products would be unchanged.

[jgoudie]

Nah, The reaction rate (number of collisions) is linked with partial pressure of each gas.  The partial pressure of each gas does not change.  Therefor no change in reaction rate.

actually the rate of the forward reaction would decrease, because there is a decrease in the number of collisions per second between reactant molecules, thus slowing the rate. However, the concentrations of the reactant and products would be unchanged.

[Limista]

Nah, The reaction rate (number of collisions) is linked with partial pressure of each gas.  The partial pressure of each gas does not change.  Therefor no change in reaction rate.

oh okay. Sorry, it's just that I came across a similar question in Cambridge Checkpoints, which stated that adding an inert gas would mean that reaction rate would decrease. Looks like they were wrong. :/ Hope the answers don't have too many more mistakes in them then - so much for the credibility of this resource!

[jgoudie]

If you could find the question i might be able to explain it.  If an inert gas is added but the total pressure held constant (ie. that means the volume must increase) this would have the rate of the reaction decrease and depending on the mole ratio in the equation the equilibrium could also shift.

However if the question just said an inert gas is added, the assumption is that the volume is held constant and that the total pressure increases, and thus nothing really happens.

Cambridge are generally quite good resources.

oh okay. Sorry, it's just that I came across a similar question in Cambridge Checkpoints, which stated that adding an inert gas would mean that reaction rate would decrease. Looks like they were wrong. :/ Hope the answers don't have too many more mistakes in them then - so much for the credibility of this resource!

[Limista]

If you could find the question i might be able to explain it.  If an inert gas is added but the total pressure held constant (ie. that means the volume must increase) this would have the rate of the reaction decrease and depending on the mole ratio in the equation the equilibrium could also shift.

However if the question just said an inert gas is added, the assumption is that the volume is held constant and that the total pressure increases, and thus nothing really happens.

Cambridge are generally quite good resources.

An important reaction in the production of ammonia, NH3, is given below.

N2(g) + 3H2(g) <--> 2NH3(g);  ΔH = -90 kJ/mol

If the reaction takes place in a sealed container which of the following procedures would not cause the rate of the forward reaction to increase?

A. Increasing the temperature
B. Increasing the pressure
C. Adding a suitable catalyst
D. Adding an inert gas

[lolipopper]

NVM

[Edward21]

An important reaction in the production of ammonia, NH3, is given below.

N2(g) + 3H2(g) <--> 2NH3(g);  ΔH = -90 kJ/mol

If the reaction takes place in a sealed container which of the following procedures would not cause the rate of the forward reaction to increase?

A. Increasing the temperature
B. Increasing the pressure
C. Adding a suitable catalyst
D. Adding an inert gas

Maybe it just means solely that the first 3 will, and D will be useless, rather than assuming it actually slows it down? Aka. D wouldn't cause the rate to increase, but wouldn't cause the rate to decrease either

[barydos]

Hey just got a little MC question that I need clarifying with.

Which of the following statements is correct with respect to an endothermic chemical reaction?
c. An increase in temperature always increases the rate of reaction.
d. The amount of energy required to break the bonds in the reactants is greater than the amount of energy released as the products form.

Thanks!

[Limista]

Maybe it just means solely that the first 3 will, and D will be useless, rather than assuming it actually slows it down? Aka. D wouldn't cause the rate to increase, but wouldn't cause the rate to decrease either

That's the problem. Apparently the reaction rate does decrease for this reaction after adding an inert gas, in spite of partial pressure stated by the collision theory. That's what I'm confused about?

Hey just got a little MC question that I need clarifying with.

Which of the following statements is correct with respect to an endothermic chemical reaction?
c. An increase in temperature always increases the rate of reaction.
d. The amount of energy required to break the bonds in the reactants is greater than the amount of energy released as the products form.

Thanks!

They are both correct, but because it's a multiple choice, I'd pick D as it is the better answer.

There are some reactions for which increasing the temperature does not increase the rate of reaction. Some reactions are virtually instantaneous - for example, a precipitation reaction involving the coming together of ions in solution to make an insoluble solid, or the reaction between hydrogen ions from an acid and hydroxide ions from an alkali in solution. So heating one of these won't make any noticeable difference to the rate of the reaction.

The number of degrees needed to double the rate in endothermic chemical reactions may change gradually as the temperature increases, to the point where the rate of reaction will no longer increase. So the doubling of rate of reaction may occur at 9 degrees, but this increase in rate of reaction may be slower at 11 degrees.

[Limista]

Having trouble with this question:

Consider the following thermochemical equations. 

C(s) + O2(g)  -->  CO2(g)                          ΔH = -393.5 kJ/mol
S(s) + O2(g)  -->  SO2(g)                          ΔH = -296.1 kJ/mol
CS2(l) + 3O2(g) -->  CO2(g) + 2SO2(g)      ΔH = -1072kJ/mol

The enthalpy of reaction (in kJ/mol) for C(s) + 2S(s) --> CS2(l) is
A. -1762
B. -873.3
C. +86.3
D. +382.4

Help is appreciated