0 votes
32 views
in ThermoChemistry by (138k points)
What is the heat evolved or absorbed when 50.0 mL of 0.100 M nitric acid (HNO3) is neutralized with 50.0 mL of 0.100 M potassium hydroxide (KOH)? Given that the standard heat of neutralization for the reaction between HNO3 and KOH is -57.1 kJ/mol.

1 Answer

0 votes
by (138k points)
To calculate the heat evolved or absorbed during the neutralization reaction, we first need to determine the number of moles of HNO3 and KOH reacting.The reaction between HNO3 and KOH can be represented as:HNO3  aq  + KOH  aq   KNO3  aq  + H2O  l Since both solutions have the same concentration and volume, they will react in a 1:1 ratio, and the limiting reactant will be completely consumed.Number of moles of HNO3 =  concentration of HNO3    volume of HNO3 = 0.100 mol/L  0.050 L= 0.005 molNumber of moles of KOH =  concentration of KOH    volume of KOH = 0.100 mol/L  0.050 L= 0.005 molNow, we can use the standard heat of neutralization to calculate the heat evolved or absorbed during the reaction.Heat evolved or absorbed =  number of moles of HNO3    standard heat of neutralization = 0.005 mol   -57.1 kJ/mol = -0.2855 kJSince the value is negative, it indicates that the heat is evolved during the reaction. Therefore, 0.2855 kJ of heat is evolved when 50.0 mL of 0.100 M nitric acid is neutralized with 50.0 mL of 0.100 M potassium hydroxide.

Related questions

Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...