A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO 3 in a coffee cup calorimeter. If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine the ΔH° rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HCl. Assume 1) that no heat is lost to the calorimeter or the surroundings, and 2) that the density and the heat capacity of the resulting solution are the same as water.

THE STANDARD HEAT OF NEUTRALIZATION OF THE BASE SODIUM HYDROXIDE BY THE ACID HYDROGEN TRIOXONITRATE V ACID IS -56 kJ / mol.

Explanation:

Volume of 0.3 M NaOh = 100 mL

Volume of 0.3 M HNO3 = 100 mL

Initail temp of NaOH and HNO3 = 35 °C = 35 + 273 K = 308 K

Final temp. of mixture = 37 °C = 37 + 273 K = 310 K

We can make the following assumptions form the question given:

1. specific heat of the reaction mixture is the same as the specific heat of water = 4.2 J/g K

2. the toal mass of the reaction mixture is 200 mL = 200 g since no heat is lost to the calorimeter or surrounding.

3. initail temperature of the reaction mixture is equal to the average temperature of the two reactant solutions

= ( 308 + 308 /2) = 308 K

4. Rise in temeperature for the reaction = 310 -308 K = 2 K

Then the total heat evolved during the reaction = mass * specifc heat capacity * temperature  change

Heat = 200 g * 4.2 J/g K * 2 K

Heat = 1680 J

EQUATION FOR THE REACTION

HNO3 + NaOH > NaNO3 + H20

From the equation, 1 mole of HNO3 reacts with 1 mole of NaOH to prouce  mole of water.

100 mL of 0.5 M HNO3 contains 100 * 0.3 /1000 = 0.03 mole of acid

This result is same for the base NaOH = 0.03 mole of base

So therefore,

0.03 mole of acid will react with 0.03 mole of base to produce 0.03 mole of water to evolved 1680 J of heat energy.

The production of 1 mole of water will evolve 1680 / 0.03 J of heat

= 56 000 J or 56 kJ of heat energy per mole of water.

So therefore, 1the standard heat of neutralization of sodium hydroxide by trioxoxnitrate V acid is -56 kJ/mol.

gas

explanation:

smoke is the matter of a gas