Question

A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculate the molar enthalpy change of the acid-base neutralization reaction to be –54 kJ/mol. He next tried the same experiment with 100 ml of 1.0 M HCl and 100 ml of 1.0 M NaOH. The calculated molar enthalpy change of reaction for his second trial was:

Answer 1

Answer: The calculated molar enthalpy change of reaction for his second trial was -108 kJ.

Explanation:-

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

`\text{no of moles}={\text{Molarity}* {\text{Volume in L}}`

Thus
`\text{no of moles}of HCl={1.0M}* {0.05L}=0.05moles`

Thus
`\text{no of moles}of NaOH={1.0M}* {0.05L}=0.05moles`

`HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)`

Given for second trial:

`\text{no of moles}of HCl={1.0M}* {0.1L}=0.1moles`

`\text{no of moles}of NaOH={1.0M}* {0.1L}=0.1moles`

0.05 moles of
`HCl` reacts with 0.05 moles of
`NaOH` to release heat = 54 kJ

0.1 moles of
`HCl` reacts with 0.05 moles of
`NaOH` to release heat =
`(54)/(0.05)* 0.1=108kJ`

Thus calculated molar enthalpy change of reaction for his second trial was -108 kJ.