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