Question

What is the value of (delta)H(reaction (in KJ/mol phosphoric acid) if 50.0ml of 0.60 M H3PO4 was used in the reaction described?

The reaction described is: A neutralzation raction was carried out in a calormeter, the change in temp was 5.6c and the mas of the solution was 100.0g. Caculate the amount of energy gained by the soulution, using 4.18J/(gxc) as the specific heat.

Answer 1

Answer:

ΔH = 89.6 kJ/mol phosphoric acid.

Energy gained: 2688 J

Explanation:

The heat (Q) of the reaction can be calculated by:

Q = m*c*ΔT

Where m is the mass of the solution, c is the specific heat, and ΔT is the change in temperature. If the energy is gained, thus Q>0, and also ΔH >0, and the reaction is endothermic.

Q = 100*4.18*5.6

Q = 2688 J

The change in enthalpy is the heat divided by the number of moles. Because it's related to the phosphoric acid, it will be the number of moles of it, which is the volume (50.0 mL = 0.05 L) multiplied by the concentration (0.60 M).

n = 0.05 * 0.60 = 0.03 mol of H3PO4

ΔH = 2688/0.03

ΔH = 89600 J/mol

ΔH = 89.6 kJ/mol phosphoric acid.

Answer 2

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

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

1. 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

2. Determination of the enthalpy change (ΔH).

We'll begin by calculating 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

Finally, we shall determine the 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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