When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 o C was measured for the beaker and its contents (Expt.1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant ( - 57.0 &#160…

Question

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of

{5.7}^{o} C

was measured for the beaker and its contents (Expt.1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant

(- 57.0 k J m o l^{- 1})

, this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid

(K_{a} = 2.0 \times 10^{- 5})

was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of

{5.6}^{o} C

was measured. (Consider heat capacity of all solutions as

4.2 J g^{- 1} K^{- 1}

and density of all solutions as

1.0 g m L^{- 1}

)
Enthalpy of dissociation

(i n k J m o l^{- 1})

of acetic acid obtained from the Expt. 2 is

Aicharya · Accepted Answer

Let the heat capacity of insulated beaker be C.
Mass of aqueous content in Expt. 1

= (100 + 100) \times 1 = 200 g

\Rightarrow T o t a l h e a t c a p a c i t y = (C + 200 \times 4.2) J / K

Moles of acid, base neutralized in Expt. 1

= 1 \times 0.1 = 0.1

\Rightarrow

Heat absorbed in Expt. 1

= 0.1 \times 57 = 5.7 k J

\Rightarrow 5.7 \times 1000 = (C + 200 \times 4.2) \times Δ T

\Rightarrow 5.7 \times 1000 = (C + 200 + 4.2) \times 5.7

\Rightarrow (C + 200 \times 4.2) = 1000

In Second experiment

n_{C H_{3} C O O H} = 0.2, n_{N a O H} = 0.1

Total mass of aqueous content = 200 g

\Rightarrow

Total heat capacity

= (C + 200 \times 4.2) = 1000

\Rightarrow

Heat released

= 1000 \times 5.6 = 5600 J

Overall, only 0.1 mol of

C H_{3} C O O H

undergo neutralization.

\Rightarrow Δ H_{n e u t r a l i z a t i o n} o f C H_{3} C O O H = \frac{- 5600}{0.1}

= - 56,000 J / m o l = - 56 K J / m o l

\Rightarrow Δ H_{i o n i z a t i o n} o f C H_{3} C O O H = 57 - 56 = 1 K J / m o l

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a…