Answer:
-54 kJ/mol
Explanation:
Given that:
A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculated the molar enthalpy change of the acid-base neutralization reaction to be –54 kJ/mol
i.e
50 ml of 1.0 M HCl + 50 ml of 1.0 M NaOH -----> -54 kJ/mol
If he repeat the same experiment with :
100 ml of 1.0 M HCl + 100 ml of 1.0 M NaOH. ------> ????
From The experiment; the molar enthalpy of change of the acid-base neutralization reaction will be -54 kJ/mol
This is because : The second reaction requires 50 ml in order to neutralize the reaction, then the remaining 50 ml will be excess, Hence, there is no change in the enthalpy of the reaction.
Similarly; we can assume that :
In the first reaction; P moles of is used to liberate Q kJ heat ; then the change in molar enthalpy will be Q/P (kJ/mol).
SO; when he used 100 ml ;
then the amount of moles used is double, likewise the heat liberated will be doubled ;
So;
2P moles is used to liberate 2Q kJ heat ;
2P/2Q = Q/P ( kJ/mol) = -54 kJ/mol
