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100 cm^3 of 0.5N HCl solution at 299.95K were mixed with 100cm^3 0.5N NaOH solution at 299.75K in a thermos flask. the final temperature was found to be 302.65K. calculate the enthalpy of neutralization of HCl. water equivalent of thermos flask is 44g.?
Most Upvoted Answer
100 cm^3 of 0.5N HCl solution at 299.95K were mixed with 100cm^3 0.5N ...
Calculation of Enthalpy of Neutralization of HCl
To calculate the enthalpy of neutralization, we need to follow these steps:
1. Calculate Heat Absorbed by the Solution
- Initial Temperatures:
- HCl: 299.95K
- NaOH: 299.75K
- Final Temperature: 302.65K
- Change in Temperature (ΔT):
- ΔT = Final Temperature - Average Initial Temperature
- Average Initial Temperature = (299.95 + 299.75) / 2 = 299.85K
- ΔT = 302.65K - 299.85K = 2.80K
- Total Volume of the Solution:
- V = 100cm³ + 100cm³ = 200cm³
- Since the density of the solution is approximately 1g/cm³, the mass of the solution (m) = 200g.
- Water Equivalent Calculation:
- Total water equivalent, W = mass of the solution + water equivalent of the flask
- W = 200g + 44g = 244g
- Heat Absorbed (q):
- q = W × specific heat capacity × ΔT
- Assuming specific heat capacity of water = 4.18 J/g°C,
- q = 244g × 4.18 J/g°C × 2.80°C = 2,835.84J
2. Calculate Moles of HCl Neutralized
- Molarity of HCl: 0.5N means 0.5 moles per liter.
- Volume of HCl = 0.1L
- Moles of HCl = 0.5 moles/L × 0.1L = 0.05 moles.
3. Calculate Enthalpy of Neutralization (ΔH)
- Enthalpy (ΔH) = -q / moles of HCl
- ΔH = -2,835.84J / 0.05 moles = -56,716.8 J/mol
Conclusion
- The enthalpy of neutralization of HCl is approximately -56.72 kJ/mol. This value indicates the exothermic nature of the neutralization reaction.
To calculate the enthalpy of neutralization, we need to follow these steps:
1. Calculate Heat Absorbed by the Solution
- Initial Temperatures:
- HCl: 299.95K
- NaOH: 299.75K
- Final Temperature: 302.65K
- Change in Temperature (ΔT):
- ΔT = Final Temperature - Average Initial Temperature
- Average Initial Temperature = (299.95 + 299.75) / 2 = 299.85K
- ΔT = 302.65K - 299.85K = 2.80K
- Total Volume of the Solution:
- V = 100cm³ + 100cm³ = 200cm³
- Since the density of the solution is approximately 1g/cm³, the mass of the solution (m) = 200g.
- Water Equivalent Calculation:
- Total water equivalent, W = mass of the solution + water equivalent of the flask
- W = 200g + 44g = 244g
- Heat Absorbed (q):
- q = W × specific heat capacity × ΔT
- Assuming specific heat capacity of water = 4.18 J/g°C,
- q = 244g × 4.18 J/g°C × 2.80°C = 2,835.84J
2. Calculate Moles of HCl Neutralized
- Molarity of HCl: 0.5N means 0.5 moles per liter.
- Volume of HCl = 0.1L
- Moles of HCl = 0.5 moles/L × 0.1L = 0.05 moles.
3. Calculate Enthalpy of Neutralization (ΔH)
- Enthalpy (ΔH) = -q / moles of HCl
- ΔH = -2,835.84J / 0.05 moles = -56,716.8 J/mol
Conclusion
- The enthalpy of neutralization of HCl is approximately -56.72 kJ/mol. This value indicates the exothermic nature of the neutralization reaction.
Community Answer
100 cm^3 of 0.5N HCl solution at 299.95K were mixed with 100cm^3 0.5N ...
The initial average temperature of the acid and the base =299.5+299.5/2=299.5 K
Rise in temperature=(302.65-299.85)=2.80 K
Heat evolved during neutralisation =(100+100+44)�4.184�2.8=2858.5 J
>>>>> Enthalpy of neutralisation= -2858.5/100�1000�1/0.50
= -57.17 kj
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100 cm^3 of 0.5N HCl solution at 299.95K were mixed with 100cm^3 0.5N NaOH solution at 299.75K in a thermos flask. the final temperature was found to be 302.65K. calculate the enthalpy of neutralization of HCl. water equivalent of thermos flask is 44g.?
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100 cm^3 of 0.5N HCl solution at 299.95K were mixed with 100cm^3 0.5N NaOH solution at 299.75K in a thermos flask. the final temperature was found to be 302.65K. calculate the enthalpy of neutralization of HCl. water equivalent of thermos flask is 44g.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 100 cm^3 of 0.5N HCl solution at 299.95K were mixed with 100cm^3 0.5N NaOH solution at 299.75K in a thermos flask. the final temperature was found to be 302.65K. calculate the enthalpy of neutralization of HCl. water equivalent of thermos flask is 44g.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 100 cm^3 of 0.5N HCl solution at 299.95K were mixed with 100cm^3 0.5N NaOH solution at 299.75K in a thermos flask. the final temperature was found to be 302.65K. calculate the enthalpy of neutralization of HCl. water equivalent of thermos flask is 44g.?.
100 cm^3 of 0.5N HCl solution at 299.95K were mixed with 100cm^3 0.5N NaOH solution at 299.75K in a thermos flask. the final temperature was found to be 302.65K. calculate the enthalpy of neutralization of HCl. water equivalent of thermos flask is 44g.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 100 cm^3 of 0.5N HCl solution at 299.95K were mixed with 100cm^3 0.5N NaOH solution at 299.75K in a thermos flask. the final temperature was found to be 302.65K. calculate the enthalpy of neutralization of HCl. water equivalent of thermos flask is 44g.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 100 cm^3 of 0.5N HCl solution at 299.95K were mixed with 100cm^3 0.5N NaOH solution at 299.75K in a thermos flask. the final temperature was found to be 302.65K. calculate the enthalpy of neutralization of HCl. water equivalent of thermos flask is 44g.?.
Solutions for 100 cm^3 of 0.5N HCl solution at 299.95K were mixed with 100cm^3 0.5N NaOH solution at 299.75K in a thermos flask. the final temperature was found to be 302.65K. calculate the enthalpy of neutralization of HCl. water equivalent of thermos flask is 44g.? in English & in Hindi are available as part of our courses for NEET.
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