Bhootnike Chem Questions

[Bhootnike]

Hi
Will list my questions for unit 4 in this thread.

1. Why don't we consider the amount of mole of both reactants, when finding the amount of energy released? e.g.
question is:

Calculate the energy released when 200 mL of diborane, B2H6, is burnt at 150°C and 1.50 atmospheres according to the following equation:

B2H6 (g) + 3 O2 (g) à B2O3 (s) + 3 H2O (g)    ∆H = -2035 kJ. mol -1

so the working in the solutions is :

pV=nRT, n = pV/RT= 8.65 x 10^-3 mol to 3sf
energy = change in h x n
= 2035 x 8.65 x 10^-3
=17.6 kJ to 3sf

so, why dont we also determine the amount of mole of 02 ? and not just b2h6 ?

2. To determine the activation energy of the reverse reaction, what 'steps' should I take?

E.G. question i found from chem dimensions:
Draw an energy profile diagram for the reaction A --> B + C for which dH = -150 kJ mol-1. The enthalpy of A can be taken as zero, and the activation energy for the forward reaction is 100 kJ mol-1.
b   Determine the activation energy of the reverse reaction B + C --> A.

Thanks a bunch.

[oneoneoneone]

Burning something implies excess oxygen, so you take the limiting reactant which is B2H6.

Activation Energy of the reverse reaction will be Activation Energy of the forward reaction - delta H.
You can see this by drawing an energy profile diagram and then looking it in in reverse.

[Bhootnike]

yeah but cause you are asked to find the amount of energy released, why don't you have to find include the number of mole of O2 that did react in the reaction ? i.e the stoichometerically equivalent amount of o2 w.r.t b2h6.

so theyre the same.. ? okli dokli!

[kenhung123]

yeah but cause you are asked to find the amount of energy released, why don't you have to find include the number of mole of O2 that did react in the reaction ? i.e the stoichometerically equivalent amount of o2 w.r.t b2h6.

so theyre the same.. ? okli dokli!

You can consider that delta H is only w.r.t. B2H6 and not combined B2H6 also

[thushan]

yeah but cause you are asked to find the amount of energy released, why don't you have to find include the number of mole of O2 that did react in the reaction ? i.e the stoichometerically equivalent amount of o2 w.r.t b2h6.

so theyre the same.. ? okli dokli!

When they say change in enthalpy, it is the change in enthalpy of the system (the mixture of reactants and products) that they're talking about. So B2H6 (g) + 3 O2 (g) <--> B2O3 (s) + 3 H2O (g)    ∆H = -2035 kJ. mol -1

This means that the difference in enthalpy between the 1 mole of B2O3 (s) + 3 moles of H2O (l) AND 1 mole of B2H6 + 3 moles of O2, is 2035 kJ, where the reactants have a higher enthalpy.

[Graphite]

Hi
Will list my questions for unit 4 in this thread.

1. Why don't we consider the amount of mole of both reactants, when finding the amount of energy released? e.g.
question is:

Calculate the energy released when 200 mL of diborane, B2H6, is burnt at 150°C and 1.50 atmospheres according to the following equation:

B2H6 (g) + 3 O2 (g) à B2O3 (s) + 3 H2O (g)    ∆H = -2035 kJ. mol -1

so the working in the solutions is :

pV=nRT, n = pV/RT= 8.65 x 10^-3 mol to 3sf
energy = change in h x n
= 2035 x 8.65 x 10^-3
=17.6 kJ to 3sf

so, why dont we also determine the amount of mole of 02 ? and not just b2h6 ?

2. To determine the activation energy of the reverse reaction, what 'steps' should I take?

E.G. question i found from chem dimensions:
Draw an energy profile diagram for the reaction A --> B + C for which dH = -150 kJ mol-1. The enthalpy of A can be taken as zero, and the activation energy for the forward reaction is 100 kJ mol-1.
b   Determine the activation energy of the reverse reaction B + C --> A.

Thanks a bunch.

1. Good question. The answer is you can if you want to. (Although I do not see a reason to make it harder for yourself)
The enthalpy of a reaction is proportional to the coefficients in the reaction.
It doesn't matter if you want to measure the amount of O2, B2H6 or combined B2H6 with O2 to calculate the energy.
What is important is that you need to recognise, is if you are calculating with O2 only, then you need to make sure the enthalpy value corresponds to the coefficient in the reaction. Likewise, if you want to use the combination of reactants, make sure the sum of the coefficients corresponds to the enthalpy value.

[Bhootnike]

To clarify, a change in the amount of species does not always imply a change in concentration of species because, if you consider adding liquid water to a system where water is a reactant:

c= n/v
if you add water, you're increasing both 'n' and 'v', hence 'c' remains constant ?

[kenhung123]

That's only because water can't really have a concentration as it is commonly the solvent which dilutes the solute.