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In a coffee-cup calorimeter, 140.0 mL of 1.1 M and 140.0 mL of 1.1 M are mixed. Both solutions were originally at 25.0°C. After the reaction, the final temperature is 32.4°C. Assuming that all the solutions have a density of 1.0 and a specific heat capacity of 4.18 J/°C ⋅ g, calculate the enthalpy change for the neutralization of by . Assume that no heat is lost to the surroundings or to the calorimeter. Enthalpy change = kJ/mol

Respuesta :

Answer:

56.23 kJ/mol

Explanation:

Given:

Total volume of the solution = 140 mL + 140 mL = 280 mL

Initial temperature of the solution = 25° C

Final temperature of the solution = 32.4° C

Density of the solution = 1.0 g/cm³

Specific heat of the solution, C = 4.18 J/°C.g

Now,

the mass of the solution = Density × Volume

or

The mass of the solution, m = 1 × 280 = 280 grams              (1 cm³ = 1 mL)

Now,

the enthalpy change = mCΔT

on substituting the respective values, we get

The enthalpy change = 280 × 4.18 × ( 32.4 - 25 )

or

The enthalpy change = 8660.96 J = 8.660 kJ

also,

the heat is evolved due to the neutralization of the solution

thus,

Number of moles = Molarity × Volume in Liter = 1.1 × ( 140 × 10⁻³ )

= 0.154 moles

Hence,

the enthalpy change for the neutralization = [tex]\frac{\textup{Enthalpy change}}{\textup{Number of moles}}[/tex]

on substituting the respective values, we get

the enthalpy change for the neutralization = [tex]\frac{8.66}{0.154}[/tex]

or

the enthalpy change for the neutralization = 56.23 kJ/mol

samueladesida43

The enthalpy change for the neutralization of the solutions is 56.23Kj/mol.

How to calculate enthalpy change of neutralization?

According to this question, 140.0 mL of 1.1 M and 140.0 mL of 1.1 M are mixed in a coffee-cup calorimeter. Both solutions were originally at 25.0°C.

Given that;

  • Total volume of the solution = 140 mL + 140 mL = 280 mL
  • Initial temperature of the solution = 25° C
  • Final temperature of the solution = 32.4° C
  • Density of the solution = 1.0 g/cm³
  • Specific heat of the solution, C = 4.18 J/°C.g

The mass of the solution = Density × Volume

The mass of the solution, m = 1 × 280 = 280 grams

Now,

Enthalpy change = mCΔT

The enthalpy change = 280 × 4.18 × ( 32.4 - 25 )

The enthalpy change = 8660.96 J = 8.660 kJ

The heat is evolved due to the neutralization of the solution

Number of moles = Molarity × Volume in Liter = 1.1 × ( 140 × 10⁻³)

= 0.154 moles

Neutralization = Enthalpy change/no of moles

Neutralization = 8.66/0.154

Neutralization = 56.23kj/mol

Therefore, the enthalpy change for the neutralization of the solutions is 56.23Kj/mol.

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