Question

g + In a coffee-cup calorimeter, when 3.25 g of NaOH is dissolved in 50.00 g of water initially at 22.0 oC, the temperature of the solution increases to 24.8 oC. Calculate the q of the reaction in kJ/mol of NaOH. Assume that the specific heat of the solution is 4.184 J/g oC, and that there is no other heat transfer than that of the solution process itself.

Answer 1

THE STANDARD HEAT OF SOLUTION OF SODIUM HYDROXIDE IN WATER IS -7.68 KJ PER MOLE.

Step-by-step 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.