Answer:
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.
