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

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 =

on substituting the respective values, we get

the enthalpy change for the neutralization =

or

the enthalpy change for the neutralization = 56.23 kJ/mol

false

because it does not have the potential to be entirely used up

the product of the pressure and volume is also constant