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. (answer + units)

Explanation:

Step1: Calculate the mass of the solution

Assume the density of the solution is \(1\space g/mL\). The total volume of the solution \(V = 100.00\space mL+50.00\space mL = 150.00\space mL\). Using the formula \(m=
ho V\), with \(
ho = 1\space g/mL\), we get \(m=(100 + 50)\space mL\times1\space g/mL=150\space g\)

Step2: Calculate the heat absorbed by the solution

Use the formula \(q = mc\Delta T\), where \(c = 4.18\space J/(g\cdot^{\circ}C)\) (specific - heat capacity of water), \(m = 150\space g\), and \(\Delta T=4.200^{\circ}C\). Then \(q=150\space g\times4.18\space J/(g\cdot^{\circ}C)\times4.200^{\circ}C\)

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Step3: Calculate the number of moles of \(HCl\)

Use the formula \(n = CV\), where \(C = 0.355\space mol/L\) and \(V=100.00\space mL=0.10000\space L\). Then \(n=0.355\space mol/L\times0.10000\space L = 0.0355\space mol\)

Step4: Calculate the molar enthalpy change

The reaction is exothermic (\(\Delta H<0\)). The formula for molar enthalpy change is \(\Delta H=-\frac{q}{n}\). Substitute \(q = 2.6334\space kJ\) and \(n = 0.0355\space mol\)

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Answer:

\(-74.2\space kJ/mol\)