When a chemical reaction is run in aqueous solution inside a calorimeter, the temperature change of the water (and Ccal) can be used to calculate the enthalpy change for the reaction. Here, we will study the reaction of hydrochloric acid with sodium hydroxide in the calorimeter from problem 3. Equal volumes (50.0 mL) of 1.00 M sodium hydroxide and 1.00 M hydrochloric acid are mixed. HCl+NaOH→NaCl+H2O
1. What is the total change in enthalpy (in Joules) for the reaction?
2. Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 j

The total change in enthalpy for the reaction is - 81533.6 J/mol

Explanation:

Given the data in the question;

Reaction;

HCl + NaOH → NaCl + H₂O

Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J

Moles of NaOH  = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

so, 0.0500 moles of H₂O produced

Volume of solution = 50.mL  +  50.mL  = 100.0 mL

Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g

now ,

Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal

we know that; The specific heat of water(H₂O) is 4.18 J/g°C.

so we substitute

q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28

q_soln = 2842.4 + 1234.28

q_soln = 4076.68 J

Enthalpy change for the neutralization is ΔH

ΔH = -q_soln / mole of water produced

so we substitute

ΔH = -( 4076.68 J ) / 0.0500 mol  

ΔH = - 81533.6 J/mol

Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol

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