What is the change in enthalpy of the following reaction?
2NaBr(aq) + PbCl2(aq) 2NaCl(aq) + PbBr2(aq)
Given:
NaCl: ∆H = -411 kJ
NaBr: ∆H = -360 kJ
PbCl2: ∆H = -359 kJ
PbBr2: ∆H = -277 kJ
Im lost with this. Everytime I try to do it, it doesn't come out right. I've taking the kJ of the products added together then did the same and subtracted the two sums and it doesn't come out right.
This equation needs reviewing. Both PbCl2 and PbBr2 are basically insoluble with very little probability of reacting as listed. The concentration of Pb^+2 is in the order of 10^-5M which translates to consuming only 2 x 10^-5M in Na-Halide. Kinetic feasibility is ~0.
Does it have to do with the signs.
Are they negative when you create the compound and positive when you break it apart?
Something for you to think about.
1st, the Standard Enthalpy values listed are for solid salts and not aqueous salts. These values can not be used in aqueous solution phase reactions.
2nd, the calculation of Enthalpy of reaction of a salt(aq) + salt(aq) => salt(aq) + ppt(s) reaction should be done using Std Heats of Formation of Ions of the respective salts. Only the ∆H(formation) of the precipitating salt in molecular form should be used.
3rd, the calculation of Enthalpy of reaction is typically done with respect to the ions and ppt driving force of the 'net ionic reaction'.
If the noted statements do not make sense to you in calculating Enthalpies of reaction of ionic pptn rxns, then further study needs to be directed toward metathesis reactions (or, double replacement reactions), their ionic equations and net ionic equations. Then, review topics and problems relating to Hess's Law of Thermal Heat flow in Chemical Reactions. Good Luck and all the best in your efforts.
To find the change in enthalpy (∆H) of a reaction, you need to apply the concept of Hess's Law. According to Hess's Law, the overall enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final states of the reaction.
In this case, we can approach the problem by using the known enthalpy changes of formation (∆Hf) for each compound involved.
1. Identify the compounds involved in the reaction and their corresponding enthalpy changes of formation:
∆Hf(NaBr) = -360 kJ/mol
∆Hf(PbCl2) = -359 kJ/mol
∆Hf(NaCl) = -411 kJ/mol
∆Hf(PbBr2) = -277 kJ/mol
2. Note that the coefficients in the balanced equation represent the number of moles of each compound.
3. Apply Hess's Law to calculate the ∆H of the reaction:
∆H = Σ (moles of product x ∆Hf(product)) - Σ (moles of reactant x ∆Hf(reactant))
Considering the balanced equation:
2NaBr(aq) + PbCl2(aq) -> 2NaCl(aq) + PbBr2(aq)
∆H = (2 mol NaCl x ∆Hf(NaCl)) + (1 mol PbBr2 x ∆Hf(PbBr2)) - (2 mol NaBr x ∆Hf(NaBr)) - (1 mol PbCl2 x ∆Hf(PbCl2))
4. Substituting the values:
∆H = (2 mol x -411 kJ/mol) + (1 mol x -277 kJ/mol) - (2 mol x -360 kJ/mol) - (1 mol x -359 kJ/mol)
5. Calculate the ∆H:
∆H = (-822 kJ) + (-277 kJ) - (-720 kJ) - (-359 kJ) = -1460 kJ
So, the change in enthalpy (∆H) of the given reaction is -1460 kJ.