Chemistry - Thermochemistry Problem Hrxn

Question

he addition of hydroiodic acid to a silver nitrate solution precipitates silver iodide according to the reaction:

AgNO3(aq)+HI(aq)→AgI(s)+HNO3(aq)

When 50.0 mL of 5.00×10⁻² M AgNO3 is combined with 50.0 mL of 5.00×10⁻² M HI in a coffee-cup calorimeter, the temperature changes from 22.40 ∘C to 22.91 ∘C.

Calculate ΔHrxn for the reaction as written. Use 1.00 g/mL as the density of the solution and Cs=4.18J/(g⋅∘C) as the specific heat capacity of the solution.

Slightly confused on how to solve this.. or how to set it up. This isn't like the other coffee cup calorimeters problems. I got it wrong on two tries...one more try left.

Answer 1

Hi Mike! Let me see if I can help.

As you identified, at its core this is a coffee-cup calorimetry problem. As such, it will obey the same equation that any other coffee-cup calorimetry problem does: q = mCΔT.

The problem directly tells us the specific heat capacity and change in temperature, but it was a bit less direct in giving us the mass of water used. What it does tell us is that 50.0 mL Silver Nitrate and 50.0 mL of Hydroiodic acid were used to perform this reaction, and that the density of the reaction solution is 1 g/mL. Working under the assumption that the volumes for this reaction are additive (that is, the volume of the product solution is the same as the total volume of the reactant solutions, which seems a fair assumption to make here), this indicates that the total volume of the solution in the coffee cup is 100.0 mL. Using the density given, this gives us 100.0 g of solution.

Now we have enough information to use the equation for coffee-cup calorimetry from above:

m = 100.0 g

C = 4.184 ^J/_g*°C

T_i = 22.40 °C

T_f = 22.91 °C

Thus, the heat energy absorbed by the solvent (water) is:

q_soln = (100.0 g)*(4.184 ^J/_g*°C)*((22.91 °C) - (22.40 °C)) = 213.384 J

Recall that our sign convention is that energy absorbed is positive, and energy released is negative. Since the water absorbed energy, the reaction itself must have released that energy. So the energy change for the reaction is q_rxn = -213.384 J.

Now there's one last thing we need to take care of. The reaction as written assumes that we combined one mole of silver nitrate with one mole of hydroiodic acid, which isn't the case in this problem. We actually used much less of the reactants than that. Recalling the math for concentrations, we see that we actually used:

(0.0500 L)*(0.0500 M) = 0.00250 mol of AgNO₃

(0.0500 L)*(0.0500 M) = 0.00250 mol of HI

Which tells us that we only used 0.00250 times what the balanced reaction tells us to. (Sometimes called "0.00250 moles of reaction.") So to get our ΔH_rxn as written, we need to divide the amount of energy released by the reaction by the amount of the reaction we actually did. That is, (-213.384 J)/(0.00250 mol_rxn) = -85353.6 J.

Thus, for the reaction as written,

ΔH_rxn = -85400 J or -85.4 kJ (rounding to three significant figures)

I hope this helps!

Tim W