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katen-ka-za [31]
3 years ago
15

A 100.00 ml of volume of 0.500 M HCl was mixed with 100.00 ml of 0.500 M KOH in a constant pressure calorimeter. The initial tem

perature of individual acid and base was 23.0 oC. The final temperature after mixing was 25.5 oC. The specific heat of the solution was 3.97 J g-1 oC-1 Assume the density of the solution is 1.00 g mL-1 a) Calculate the heat of the reaction in kJ. (Be sure to use signs correctly.) b) Calculate the enthalpy of the reaction in kJ/mol. (Be sure to use signs correctly.) DO NOT put units in the answer box or your answer will be counted wrong.
Chemistry
1 answer:
Virty [35]3 years ago
7 0

Answer:

The heat of the reaction = -1985 J = -1.985 kJ

The enthalpy of the reaction is -39.7 kJ/ mol

Explanation:

<u>Step 1: </u>Data given

Volume of HCl = 100 mL the heat of the reaction = 0.1 L

Molarity of HCl solution = 0.500 M

Volume of KOH = 100 mL = 0.1 L

Molarity of KOH solution = 0.500 M

Initial temperature = 23.0 °C

Final temperature = 25.5 °C

Specific heat of the solution = 3.97 J/°C *g

Density of the solution = 1g/ mL

<u>Step 2: </u>Calculate heat

Q = m*c*ΔT

with m = the mass of both solution : 100g + 100 g ( since density = 1g/mL) = 200 g

c = the specific heat = 3.97 J/°C*g

ΔT  = T2 -T1 = 25.5 = 23 = 2.5 °C

Q = 200g *3.97 J/°C*g * 2.5°C = 1985 J  (= -1985 J because it's exothermic)

<u />

<u>Step 3:</u> Calculate the number of moles

Number of moles = Molarity * Volume

Number of moles = 0.5 * 0.1 L = 0.05 moles

(Moles of the acid are equal to the moles of water produced.

Moles of solution = 0.05 moles)

<u>Step 4: </u>Calculate the enthalpy of the reaction

ΔH = heat change /Number of moles

    = -1.985 kJ/ 0.05 moles

   =- 39700 J/mol = -39.7 kJ/ mol

The enthalpy of the reaction is -39.7 kJ/ mol

The enthalpy of the reaction is negative, this means the reaction is exothermic ( which means the final temperature is higher than the initial temperature.)

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