Answer:
THE STANDARD HEAT OF SOLUTION OF SODIUM HYDROXIDE IN WATER IS -7.68 KJ PER MOLE.
Explanation:
Variables:
Mass of NaOH = 3.25 g
Mass of water = 50 g
Initial temperature of water = 22°C = 22 + 273 K = 295 K
Final temperature of the reaction mixture = 24.8 °C = 24.8 + 273 K = 297.8 K
Assuming that:
1. specific heat of water = 4.184 J/g °C
2. total mass of the reaction mixture = 50 g + 3.25 g = 53.25 g
3. the rise in temperature = (297.8 K - 295 K ) = 2.8 K
4. Molar mass of sodium hydroxide = ( 23 + 16 + 1) = 40 g/mol
5. number of mole of sodium hydroxide = mass / molar mass
n = 3.25 g / 40 g/mol
n = 0.08125 moles
The rise in temperature for the reaction mixture produces how much of heat:
Heat = mass * specific heat * change in temperature
Heat = 53.25 * 4.184 * 2.8
Heat = 623.8344 J of heat.
Equation of reaction:
NaOH + H2O -------> NaOH + H2O + Heat
This is not a reaction but a dissolution as sodium hydroxide is very soluble in water and this reaction is exothermic where heat is given off.
So since 3.25 g having 0.08125 moles produces 623.8344 J of heat, 1 mole of the sodium hydroxide used will produce:
0.08125 mole of sodium hydroxide = 623.8344 J of heat
1 mole of sodium hydroxide = ( 623.8344 / 0.08125 J of heat
= 7677.96 J of heat per mole of sodium hydroxide.
= 7.68 kJ of heat
So therefore, the standard heat of solution of sodium hydroxide in water is -7.68 kJ of heat since its an exothermic reaction.
