Enthelpy of reaction calculation?

[Centralscience]

Limestone stalactites and stalagmites are formed in caves by the following reaction:
Ca2+(aq) + 2HCO3-(aq) --> CaCO3(s) + CO2(g) + H2O(l) . If one mol of CaCO3 forms at 298 K under 1 atm pressure, the reaction performs 2.48 kJ of P - V work, pushing back the atmosphere as the gaseous CO2 forms. At the same time, 38.95 kJ of heat is absorbed from the environment.
A.) What is the value of ΔH for this reaction?
B.) What is the value of ΔE for this reaction?

So I thought that I had to add together the standard enthalpies of formation for each of the products and subtract the standard enthalpies of formation for the reactants. However, I did not see any ΔHf for Ca2+ or HCO3-. Secondly , when I looked at the answers, it said that the answer was ΔH=38.95, ΔE=36.47.

I guess I don't understand when we should use the standard enthalpies of formation. Also, do ions made up of a single element have enthalpies of formation? And if I did have all the necessary enthalpies of formation, would my answer for ΔH also be 38.95?

Thanks!

 

[Borek]

You can think about ΔE and ΔH as if they were amounts of energy exchanged with the surroundings (just in different conditions). As the energy is conserved whatever is lost/gained by the reagents must be gained/lost by the world around.

 

[Chestermiller]

Limestone stalactites and stalagmites are formed in caves by the following reaction:
Ca2+(aq) + 2HCO3-(aq) --> CaCO3(s) + CO2(g) + H2O(l) . If one mol of CaCO3 forms at 298 K under 1 atm pressure, the reaction performs 2.48 kJ of P - V work, pushing back the atmosphere as the gaseous CO2 forms. At the same time, 38.95 kJ of heat is absorbed from the environment.
A.) What is the value of ΔH for this reaction?
B.) What is the value of ΔE for this reaction?

So I thought that I had to add together the standard enthalpies of formation for each of the products and subtract the standard enthalpies of formation for the reactants. However, I did not see any ΔHf for Ca2+ or HCO3-. Secondly , when I looked at the answers, it said that the answer was ΔH=38.95, ΔE=36.47.

I guess I don't understand when we should use the standard enthalpies of formation. Also, do ions made up of a single element have enthalpies of formation? And if I did have all the necessary enthalpies of formation, would my answer for ΔH also be 38.95?

Thanks!

The implication of this problem statement is that exactly the right amount of heat is added so that the temperature (298) doesn't change. In this case, the reaction is carried out at a constant pressure of 1 atm and a constant temperature of 298 K. Therefore, from the definition of the heat of reaction, what is the ΔH for this reaction? You also know that Δ(PV) = 2.48 kJ. From the equation, ΔH = ΔE +Δ(PV), what is ΔE?

There is no need to consider heats of formation, because the heat of the reaction is already given.

Chet

 

[gracy]

, because the heat of the reaction is already given.

38.95 kJ of heat?

 

[Chestermiller]

38.95 kJ of heat?

Sure.

Chet

 

[gracy]

enthalpy of the reaction is denoted by ΔH ,right?

 

[Chestermiller]

In the present context, ΔH is defined as the enthalpy of the products minus the enthalpy of the reactants, starting out with molar stoichiometric quantities of reactants. It is equal to the amount of heat Q that has to be added to the system so that the final temperature of the products is equal to the temperature of the reactants.

Chet