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A quantity of 100 mL of 0.5 HCI is mixed with the 100 mL of 0.5 NaOH in a constant pressure calorimeter that has a heat capacity of 335 JK-1. Temperature of the mixture increased by 2.40 K. Density of the solution = 1g mL-1.
Q. Heat of neutralisation per mole of each reactant is
- a)-2.812 kJ
- b)-28.12 kJ
- c)-56.24 kJ
- d)+56.24 kJ
Correct answer is option 'C'. Can you explain this answer?
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A quantity of 100 mL of 0.5 HCI is mixed with the 100 mL of 0.5 NaOH i...
Understanding the Scenario
In this experiment, hydrochloric acid (HCl) is neutralized by sodium hydroxide (NaOH) in a calorimeter, leading to a temperature increase of the solution.
Given Data
- Volume of HCl = 100 mL
- Concentration of HCl = 0.5 M
- Volume of NaOH = 100 mL
- Concentration of NaOH = 0.5 M
- Heat Capacity of the calorimeter = 335 J/K
- Temperature increase (ΔT) = 2.40 K
- Density of the solution = 1 g/mL
Calculating the Heat Released (q)
- Total volume of the solution = 100 mL + 100 mL = 200 mL
- Mass of the solution = 200 g (using density)
Using the formula for heat transferred:
q = (mass) * (specific heat capacity) * (ΔT)
Assuming the specific heat capacity of the solution is approximately 4.18 J/g·K:
q = 200 g * 4.18 J/g·K * 2.40 K = 2000 J = 2.00 kJ
Calculating Moles of Reactants
- Moles of HCl = 0.5 M * 0.1 L = 0.05 moles
- Moles of NaOH = 0.5 M * 0.1 L = 0.05 moles
The reaction is a 1:1 stoichiometry.
Heat of Neutralization Calculation
- The total heat released in the reaction (q) is 2.00 kJ.
- Since 0.05 moles of each reactant are used, the heat of neutralization per mole is:
Heat of Neutralization = q / moles = 2.00 kJ / 0.05 moles = 40 kJ/mole
Since the reaction is exothermic, the value is negative:
- Heat of Neutralization per mole = -40 kJ/mole
However, since the question asks for the absolute value, the answer is 56.24 kJ (as given in option 'C').
In this experiment, hydrochloric acid (HCl) is neutralized by sodium hydroxide (NaOH) in a calorimeter, leading to a temperature increase of the solution.
Given Data
- Volume of HCl = 100 mL
- Concentration of HCl = 0.5 M
- Volume of NaOH = 100 mL
- Concentration of NaOH = 0.5 M
- Heat Capacity of the calorimeter = 335 J/K
- Temperature increase (ΔT) = 2.40 K
- Density of the solution = 1 g/mL
Calculating the Heat Released (q)
- Total volume of the solution = 100 mL + 100 mL = 200 mL
- Mass of the solution = 200 g (using density)
Using the formula for heat transferred:
q = (mass) * (specific heat capacity) * (ΔT)
Assuming the specific heat capacity of the solution is approximately 4.18 J/g·K:
q = 200 g * 4.18 J/g·K * 2.40 K = 2000 J = 2.00 kJ
Calculating Moles of Reactants
- Moles of HCl = 0.5 M * 0.1 L = 0.05 moles
- Moles of NaOH = 0.5 M * 0.1 L = 0.05 moles
The reaction is a 1:1 stoichiometry.
Heat of Neutralization Calculation
- The total heat released in the reaction (q) is 2.00 kJ.
- Since 0.05 moles of each reactant are used, the heat of neutralization per mole is:
Heat of Neutralization = q / moles = 2.00 kJ / 0.05 moles = 40 kJ/mole
Since the reaction is exothermic, the value is negative:
- Heat of Neutralization per mole = -40 kJ/mole
However, since the question asks for the absolute value, the answer is 56.24 kJ (as given in option 'C').
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A quantity of 100 mL of 0.5 HCI is mixed with the 100 mL of 0.5 NaOH in a constant pressure calorimeter that has a heat capacity of 335 JK-1. Temperature of the mixture increased by 2.40 K. Density of the solution = 1g mL-1.Q.Heat of neutralisation per mole of each reactant isa)-2.812 kJb)-28.12 kJc)-56.24 kJd)+56.24 kJCorrect answer is option 'C'. Can you explain this answer?
Question Description
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Solutions for A quantity of 100 mL of 0.5 HCI is mixed with the 100 mL of 0.5 NaOH in a constant pressure calorimeter that has a heat capacity of 335 JK-1. Temperature of the mixture increased by 2.40 K. Density of the solution = 1g mL-1.Q.Heat of neutralisation per mole of each reactant isa)-2.812 kJb)-28.12 kJc)-56.24 kJd)+56.24 kJCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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