JEE Advance - Chemistry (2015 - Paper 2 Offline - No. 6)
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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 5.7o 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 constant (-57.0 kJ/mol), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2) 100 mL of 2.0 M acetic acid (Ka = 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 5.6o C was measured. (Consider heat capacity of all solutions as 4.2 J/gK and density of all solutions as 1.0 g m/L)
Question
Enthalpy of dissociation (in kJ/mol) of acetic acid obtained from the Expt. 2 is
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 5.7o 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 constant (-57.0 kJ/mol), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2) 100 mL of 2.0 M acetic acid (Ka = 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 5.6o C was measured. (Consider heat capacity of all solutions as 4.2 J/gK and density of all solutions as 1.0 g m/L)
Question
Enthalpy of dissociation (in kJ/mol) of acetic acid obtained from the Expt. 2 is
1.0
10.0
24.5
51.4
Explanation
Energy evolved on neutralization of HCl and NaOH is
0.1 $$\times$$ 57 = 5.7 kJ = 5700 J
Energy utilized to rise the temperature of the solution is
ms . $$\Delta$$T = 200 $$\times$$ 1 $$\times$$ 4.2 $$\times$$ 5.7 = 4788 J
Energy used to increase temperature of calorimeter is
= 5700 $$-$$ 4788 = 912 J
ms . $$\Delta$$T = 912
m $$\times$$ s $$\times$$ 5.7 = 912 $$\rightarrow$$ ms = 160 J$$^\circ$$C$$-$$1 [Calorimeter constant]
Energy evolved by neutralization of CH3COOH and NaOH is
= 200 $$\times$$ 4.2 $$\times$$ 5.6 + 160 $$\times$$ 5.6 = 5600 J
So, the energy used in dissociation of 0.1 mol CH3COOH is
= 5700 $$-$$ 5600 = 100 J
Thus, enthalpy of dissociation is 1 kJ mol$$-$$1.
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