A calorimeter contains 26.0 mL of water at 12.5 C . When 1.50 g of X (a substanc

Question

A calorimeter contains 26.0 mL of water at 12.5 C . When 1.50 g of X (a substance with a molar mass of 76.0 g/mol ) is added, it dissolves via the reaction X(s)+H 2 O(l)X(aq) and the temperature of the solution increases to 28.0 C .

Calculate the enthalpy change, H , 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. kJ/mol please!

Explanation / Answer

Moles of X = 1.50/76 = 0.0197 moles

Since there is a increase in temperature of the solution

We can say that the solution will gain heat from the solid ! And the solid will be libearting heat and the solution will be gaining which is our required delta H or the enthalpy

= m*S* delta T = delta H

= 26 grams * 4.18*J/(grams*C) * (28-12.5)*C = delta H

= 1684.54 Joules = delta H

so know as we need for per mole

This values is for 0.0197 moles of X

so for 1 mole = 1684.54 /0.0197 = 85509.64 J/mole

Now we need the ans in KJ/mol = 85.509 KJ/mole

3 significant figure = 85.5 KJ/mole

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