Use the chemical reaction below to answer the following questions HCl(aq) + NaOH(aq) → NaCl (aq) + H2O (I) AH298 = -58 kJ/mol (a) How much heat is produced when 200 mL of 0.3 M HCL (density = 1.00 g/mL) and 250 mL of 0.25 M NaOH (density = 1.00 g/mL) are mixed? (b) If both of the solutions at the same temperature and the heat capacity of the products are 5.39 J/g °C , how much will the temperature increase? (c) What assumptions are made when calculating your answers
Use the chemical reaction below to answer the following questions HCl(aq) + NaOH(aq) → NaCl (aq) + H2O (I) AH298 = -58 kJ/mol (a) How much heat is produced when 200 mL of 0.3 M HCL (density = 1.00 g/mL) and 250 mL of 0.25 M NaOH (density = 1.00 g/mL) are mixed? (b) If both of the solutions at the same temperature and the heat capacity of the products are 5.39 J/g °C , how much will the temperature increase? (c) What assumptions are made when calculating your answers
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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![### Chemical Reaction Analysis
Use the chemical reaction below to answer the following questions:
\[ \text{HCl(aq) + NaOH(aq) → NaCl(aq) + H}_2\text{O (l)} \hspace{10pt} \Delta H_{298}^\circ = -58 \text{ kJ/mol} \]
#### Questions:
(a) **How much heat is produced when 200 mL of 0.3 M HCl (density = 1.00 g/mL) and 250 mL of 0.25 M NaOH (density = 1.00 g/mL) are mixed?**
(b) **If both of the solutions are at the same temperature and the heat capacity of the products is 5.39 J/g°C, how much will the temperature increase?**
(c) **What assumptions are made when calculating your answers?**
### Explanation:
- **Reaction Details:** The reaction is an exothermic neutralization reaction, meaning it releases heat. The enthalpy change (\(\Delta H_{298}^\circ\)) is -58 kJ/mol, indicating the amount of heat released per mole of reaction.
- **Calculating Heat Production:** To find out how much heat is produced, you'll need to calculate the moles of each reactant and determine the limiting reactant. Use the stoichiometric relationship to find the heat released based on the amount of limiting reactant.
- **Temperature Increase:** To find how much the temperature will increase, use the formula involving specific heat capacity:
\[ q = mc\Delta T \]
Solve for \(\Delta T\), where \(q\) is the heat absorbed (opposite of heat released), \(m\) is the mass of the solution, and \(c\) is the specific heat capacity.
- **Assumptions:** These may include ideal behavior of the solutions, complete reaction, no heat loss to the surroundings, and uniform specific heat capacity for the solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F73303aca-d795-4569-9e8f-a8984c4a41f6%2Fcecad1f0-b813-4679-a1cf-980f443c07b1%2Fhz2az8o_processed.png&w=3840&q=75)
Transcribed Image Text:### Chemical Reaction Analysis
Use the chemical reaction below to answer the following questions:
\[ \text{HCl(aq) + NaOH(aq) → NaCl(aq) + H}_2\text{O (l)} \hspace{10pt} \Delta H_{298}^\circ = -58 \text{ kJ/mol} \]
#### Questions:
(a) **How much heat is produced when 200 mL of 0.3 M HCl (density = 1.00 g/mL) and 250 mL of 0.25 M NaOH (density = 1.00 g/mL) are mixed?**
(b) **If both of the solutions are at the same temperature and the heat capacity of the products is 5.39 J/g°C, how much will the temperature increase?**
(c) **What assumptions are made when calculating your answers?**
### Explanation:
- **Reaction Details:** The reaction is an exothermic neutralization reaction, meaning it releases heat. The enthalpy change (\(\Delta H_{298}^\circ\)) is -58 kJ/mol, indicating the amount of heat released per mole of reaction.
- **Calculating Heat Production:** To find out how much heat is produced, you'll need to calculate the moles of each reactant and determine the limiting reactant. Use the stoichiometric relationship to find the heat released based on the amount of limiting reactant.
- **Temperature Increase:** To find how much the temperature will increase, use the formula involving specific heat capacity:
\[ q = mc\Delta T \]
Solve for \(\Delta T\), where \(q\) is the heat absorbed (opposite of heat released), \(m\) is the mass of the solution, and \(c\) is the specific heat capacity.
- **Assumptions:** These may include ideal behavior of the solutions, complete reaction, no heat loss to the surroundings, and uniform specific heat capacity for the solution.
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