Question

When a 4.98 g of NaOH was dissolved in 49.72g of water in a calorimeter at 23.7 c, the temperature of the solution went up to 50.1 c

1.) calculate qh2o

2.) find delta H for the reaction as itoccured in the callorimeter using this equation: q reaction= delta h reaction = qh20

3.) find delta h for the dissolution of 1.00 g NaOH in water

4) find delta H for the dissolution of 1 mole in NaOH in water

5) given that NAaOh exists as Na+ and OH- ions in solution, write the equation for the reaction that occurs when NaOH is dissolved in water.

Answer 1

1) To calculate qH2O (heat gained by water), we can use the formula:
qH2O = mass of water × specific heat of water × change in temperature

The specific heat of water is 4.184 J/g°C, the mass of water is 49.72 g, and the change in temperature is (50.1 - 23.7) °C.

qH2O = 49.72 g × 4.184 J/g°C × (50.1 - 23.7) °C
qH2O = 49.72 g × 4.184 J/g°C × 26.4 °C
qH2O = 5470.73 J

2) Since the heat gained by water (qH2O) is equal to the heat released by the reaction (q reaction), we have:
ΔH reaction = qH2O = 5470.73 J

3) To find ΔH for the dissolution of 1.00 g NaOH in water, we need to divide the ΔH reaction by the mass of NaOH used in the experiment:
ΔH for 1 g NaOH = ΔH reaction / mass of NaOH
ΔH for 1 g NaOH = 5470.73 J / 4.98 g
ΔH for 1 g NaOH = 1098.14 J/g

4) To find ΔH for the dissolution of 1 mole NaOH in water, we need to multiply the ΔH for 1 g NaOH by the molar mass of NaOH (22.99 g/mol for Na, 15.999 g/mol for O, and 1.007 g/mol for H):
Molar mass of NaOH = 22.99 + 15.999 + 1.007 = 39.996 g/mol

ΔH for 1 mole NaOH = ΔH for 1 g NaOH × molar mass of NaOH
ΔH for 1 mole NaOH = 1098.14 J/g × 39.996 g/mol
ΔH for 1 mole NaOH = 43947.93 J/mol

5) The equation for the reaction that occurs when NaOH is dissolved in water can be written as:
NaOH (s) → Na⁺ (aq) + OH⁻ (aq)