A100.00 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 temperature 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.

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

The enthalpy of the reaction is -39.7 kJ/ mol

Explanation:

Step 1: 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

Step 2: 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)

Step 3: 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)

Step 4: 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.)

the ratio of the conjugate base to the acid required to attain the desired ph.

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