Question

2. A neutralization reaction was carried out in a calorimeter. The change in temperature (AT) of the solution was 5.6 °C and the mass of the solution was 100.0 g. Calculate the amount of heat energy gained by the solution (qsol). Use 4.18 J/(g.C) as the specific heat, Cs, of the solution.

3. What is the value of qreaction for the neutralization reaction described in number 2?

4. How many moles of phosphoric acid are contained in 50.0 mL of 0.60 M H3PO4?

5. What is the value of AHr kJ/mol phosphoric acid) if 50.0 mL of 0.60 M reaction in H3PO4 was used in the reaction described in number 2?

Answer 1

Explanation:

2. First we have to calculate the heat gained by the solution.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat of solution = $5.20kJ/^oC$

m = Mass of the solution = 100.0 g

$T_{final}$ = final temperature

$T_{initial}$ = initial temperature

ΔT = $T_{final}-T_{initial}$ = 5.6°C

Now put all the given values in the above formula, we get:

$q_{solution}=100.0 g\times 4.18 J/^oC\times (5.6)^oC$

$q_{solution}=2,340.8 J$

The amount of heat energy gained by the solution 2,340.8 J.

3.

Heat gained by solution = Heat released on neutralization reaction

$q_{solution}=q_{recation}=2,340.8 J$

2,340.8 Joules is the value $q_{reaction }$ of reaction for the neutralization reaction described in number 2.

4. $Molarity=\frac{Moles}{Volume(L)}$

Moles of phosphoric acid = n

Volume of the phosphoric acid solution = 50.0 mL = 0.050 L

Molarity of the phosphoric acid solution = 0.60 M

$n=0.60 M\times 0.050 L= 0.03 mol$

0.03 moles of phosphoric acid are contained in 50.0 mL of 0.60 M phosphoric acid solution.

5. To calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

n = moles of phosphoric acid = 0.03 moles

$\Delta H$ = enthalpy change = ?

q = heat gained = 2,340.8 J = 2.3408 kJ

$\Delta H=-\frac{ 2.3408 kJ}{0.03 mol}=-78.027 kJ/mole$

Therefore, the enthalpy change during the reaction is -78.027 kJ/mole

Answer 2

2. The amount of heat gained by the reaction is 2340.8 J

3. The value of heat (q) for the reaction is –2340.8 J

4. The number of mole of H₃PO₄ in the solution is 0.03 mole

5. The value of enthalpy change, ΔH for the reaction is 78.03 KJ/mol

2. Determination of the heat gained by the solution.

• Mass (M) = 100 g

• Change in temperature (ΔT) = 5.6 °C

• Specific heat capacity (C) = 4.18 J/gºC

• Heat gained (Q) =?

Q = MCΔT

Q = 100 × 4.18 × 5.6

Q = 2340.8 J

3. Determination of the heat (q) of the reaction.

• Heat gained = 2340.8 J

• Heat of reaction =?

Heat released = –heat gained

Heat released = –2340.8 J

Thus, the heat for the reaction is –2340.8 J

4. Determination of the number of mole of H₃PO₄ in the solution

• Volume = 50 mL = 50 / 1000 = 0.05 L

• Molarity = 0.6 M

• Mole of H₃PO₄ =?

Mole = Molarity x Volume

Mole of H₃PO₄ = 0.6 × 0.05

Mole of H₃PO₄ = 0.03 mole

5. Determination of the enthalpy change, ΔH

• Mole of H₃PO₄ = 0.03 mole

• Energy (Q)  = 2340.8 J = 2340.8 /1000 = 2.3408 KJ

• Enthalpy change (ΔH) =?

ΔH = Q/n

ΔH = 2.3408 / 0.03

ΔH = 78.03 KJ/mol

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