Question

1. hcl is a corrosive colourless gas that dissolves readily in water. aqueous hcl reacts with naoh to form water and nacl. in a simple calorimeter, a 100.00 ml sample of 0.355 mol/l hcl(aq) is mixed with 50.00 ml of excess naoh(aq). during the reaction, there is a rise in temperature by 4.200 °c. calculate the molar enthalpy change for the above reaction. show all your work.

Explanation:

Step1: Calculate the moles of HCl

The formula for moles \(n = c\times V\), where \(c\) is the concentration and \(V\) is the volume.
Given \(c = 0.355\space mol/L\) and \(V=100.00\space mL=0.10000\space L\)
\(n_{HCl}=0.355\space mol/L\times0.10000\space L = 0.0355\space mol\)

Step2: Calculate the mass of the solution

Assume the density of the solution is \(1\space g/mL\). The total volume of the solution \(V_{total}=100.00 + 50.00=150.00\space mL\)
The mass \(m=
ho\times V\), with \(
ho = 1\space g/mL\) and \(V = 150.00\space mL\)
\(m = 150.00\space g\)

Step3: Calculate the heat absorbed by the solution

The formula for heat \(q = mc\Delta T\), where \(c = 4.18\space J/(g\cdot^{\circ}C)\) (specific - heat capacity of water) and \(\Delta T=4.200^{\circ}C\)
\(q=(150.00\space g)\times(4.18\space J/(g\cdot^{\circ}C))\times(4.200^{\circ}C)\)
\(q = 150\times4.18\times4.200\space J=2633.4\space J = 2.6334\space kJ\)

Step4: Calculate the molar enthalpy change

The molar enthalpy change \(\Delta H=\frac{q}{n}\)
Since the reaction is exothermic (\(q\) is released by the reaction), \(\Delta H=-\frac{2.6334\space kJ}{0.0355\space mol}\)
\(\Delta H=- 74.2\space kJ/mol\)

Answer:

The molar enthalpy change for the reaction is \(-74.2\space kJ/mol\)