Answer to Question #159073 in General Chemistry for Arvin Shopon

Question

4. Calculate the enthalpy of the combustion reaction of methanol using the binding energies.

5. In a calorimeter, a student dissolved 4.25 g of solid ammonium nitrate in 60.0 ml of water with an initial temperature of 22.0 °C. When the ammonium nitrate was completely dissolved, the temperature of the solution was 16.9 °C. Based on the measurement results, calculate the heat of dissolution of ammonium nitrate in kJ/mol.

Answer 1

Q159073

4. Calculate the enthalpy of the combustion reaction of methanol using the binding energies.

Solution:

The combustion of methanol is given as

2CH₃OH(g) + 3 O₂ (g) ==> 2 CO₂ (g) + 4 H₂O (g)

The calculated heat of combustion will vary depending on the data provided to us.

I have used the data given in the table and calculated

The heat of combustion of methanol = -1340 kJ

In the reaction the coefficient of CH₃ OH is 2.

So enthalpy of combustion for 1 mole of methanol will be = -1340 kJ/2 mol = -670kJ/mol .

Enthalpy for combustion reaction will be = -1340 kJ

and enthalpy of reaction per mole of methanol will be - 670 kJ/mol.

5. In a calorimeter, a student dissolved 4.25 g of solid ammonium nitrate in 60.0 ml of water with an initial temperature of 22.0 °C. When the ammonium nitrate was completely dissolved, the temperature of the solution was 16.9 °C. Based on the measurement results, calculate the heat of dissolution of ammonium nitrate in kJ/mol.

Solution :

Let us assume that the density of water is 1.00g/mL

So the mass of 60.0 mL of water will be equal = 60.0 grams.

4.25g of NH₄NO₃ is added to 60 grams of water.

So mass of solution = 4.25g + 60.0 g = 64.25 grams.

Step 1: To find the heat given by the dissolution of ammonium nitrate.

We will use the specific heat capacity formula

Q = m * s *  ∆T = 64.25 g * 4.184 J/g ⁰C * ( 16.9 0C - 22.0 ⁰C)

Q = 64.25 * 4.184 J/g ⁰C * (-5.1 ⁰C )

Q = -1370.99 Joules.

Step 2: To find Heat of dissolution of ammonium nitrate in kJ/mol

grams of NH₄NO3 = 4.25 grams

molar mass of NH₄NO₃ = 2 * atomic mass of N + 4 * atomic mass of H + 3 * atomic mass of O

= 2 * 14.007 g/mol + 4 * 1.00794 g/mol + 3 * 15.999 g/mol

= 28.014g/mol + 4.03176 g/mol +47.997 g/mol

= 80.043g/mol

moles of NH₄NO₃ = 4.25 g NH₄NO₃ * 1mol NH₄NO₃ / 80.043 g NH₄ NO₃

= 0.05309 moles

Heat of dissolution of ammonium nitrate = -1370.99 Joules / 0.05309 mol

= -25820 Joules/mol

Convert this to kilojoules using 1kilojoule = 1000 Joule

in kilojoule the answer will be = -25820 J/mol * 1kJ/ 1000 J

= -258 kilojoules per mole

The heat of dissolution of ammonium nitrate is -258 kilojoules per mole