Question

Nitric acid neutralizes potassium hydroxide. To determine the heat of reaction, a student placed 56.6 mL of 1.3 M HNO3 in a coffee cup calorimeter, noted that the temperature was 22.8 °C, and added 56.6 mL of 1.3 M KOH, also at 22.8 °C. The mixture was stirred quickly with a thermometer, and its temperature rose to 36.2 °C.

Balance equation for the reaction. Do not include physical states.

Calculate the heat of reaction in joules. Assume that the specific heats of all solutions are 4.18 J·g^-1·°C^-1 and that all densities are 1.00 g·mL^-1.

Calculate the heat of reaction per mole of acid (in units of kJ·mol^-1).

Answer 1

temperature before the reaction =22.8 deg.c

moles of KOH= moles of HNO3= molarity* volume in L= 1.3*56.6/1000 =0.074

the reaction is HNO3+ KOH ----->KNO3+ H2O

1 mole of KNO3 reacts with 1 mole of KOH to produce 1 mole of KNO3.

volume of KNO3= volume of KOH= 56.6 ml, total volume of mixture= 56.6+56.6= 113.2 ml

mass of the solution = volume of solution*density = 113.2ml*1g/ml =113.2 gm

temperature after the reaction = 36.2 deg.c, temperature rise during the reaction= 36.2-22.8=13.4 deg.c

enthalpy of reaction= mass of solution* specific heat* temperature difference= 113.2*4.184*13.4 Joules=6347 joules

since moles of KOH= moles of HNO3, all the moles of KOH as well as HNO3 react since as per the reaction, ratio of HNO3: KOH= 1:1, enthalpy of reactopn= heat liberated/ moles=6347/0.074 j/mole =85770J/mole= 85770/1000 Kj/mole= 85.770 KJ/mole