Determine the ΔrH and ΔrU for the reaction

Question

Magnesium metal reacts with acid in solution according to the following reaction:

Mg(s)+2H+(aq)→Mg2+(aq)+H2(g)

When 0.1021 g of Mg(s) is combined with enough acid solution to make 150.0 g of solution in a coffee-cup calorimeter, all of the magnesium reacts, raising the temperature from 21.80 ∘C to 24.90 ∘C. Assume that the solution is mostly water, so that the heat capacity is 4.184 J⋅g−1⋅∘C−1.

1) Determine ΔrH for the reaction (at 298 K).

2)Determine ΔrU for the reaction (at 298 K)

Hi, I am struggling to solve this question. Please help me solve it. Thanks.

Answer 1

Considering that the pressure and the volume are constant, ∆H = q and ∆U = ∆H

q = mC∆T

q = heat = ?

m = mass = 150.0 g

C = specific heat = 4.184 J/gº

∆T = change in temperature = 24.90º - 21.80º = 3.10º

q = (150.0 g)(4.184 J/gº)(3.10º) = -1946 J

moles Mg = 0.1021 g Mg x 1 mol / 24.31 g = 0.004200 moles

q_rxn = -1946 J / 0.0042 mols = -463,333 J/mol = -463.3 kJ/mol = ∆Hrxn (∆rH) = ∆U (∆rU)