A calorimeter contains 19.0mL of water at 15.0∘C . When 1.70g of X (a substance with a molar mass of 79.0g/mol ) is added, it dissolves via the reaction

X(s)+H2O(l)→X(aq)
and the temperature of the solution increases to 26.0∘C .

Calculate the enthalpy change, ΔH, for this reaction per mole of X.

Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g⋅∘C)], that density of water is 1.00 g/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings.
Express the change in enthalpy in kilojoules per mole to three significant figures.

do you have a beginning temperature of the substance? if not, i guess we will just assume the starting temperature of X is 15 degrees Celsius as well.

So the resulting mass of the solution is calculated as follows:
19 ml h2o @ 1 g/ml = 19 g
mass of X added = 1.7 g

Total mass = 20.7 g

Temperature change: 26-15 = 11.

heat change for solution = 4.18 j/(g*C) * 20.7 g *11 C = 951.786 J
heat change for environment = - heat change for solution = -951.786 J = -.951786 kJ

Now calculate mol of X added : 1.7/79 = 0.0215189873417722

delta H = -.951786kJ/0.0215189873417722 mol = -44.23 kJ/mol

believe this should be right.

very appreciated man, my mess up was that i had a positive answer rather then a negative. Thank you again!

One last question that I cant figure out



Consider the reaction

2Al(OH)3(s)→Al2O3(s)+3H2O(l)
with enthalpy of reaction

ΔHrxn∘=21.00kJ/mol
What is the enthalpy of formation of Al2O3(s) ?
Express your answer in kilojoules per mole to four significant figures.


Substance ΔHf∘ (kJ/mol)
HCl(g) −92
Al(OH)3(s) −1277
H2O(l) −285.8
AlCl3(s) −705.6
H2O(g) −241.8

So for Rxn you have

Rxn = sum of enthalpy of formation of products - sum of enthalpy of formation of reactants

Remember if 2 mol of Product, take that into consideration

Was just about to edit post saying I figured it out. I just messed up the last step. Thank you very much though

Np