Question

<p>"When a 6.50-g sample of solid sodium hydroxide disolves in 100.0 g of water in a coffee-cup calorimeter, the temperature rises form 21.6 degrees Celcius to37.8 degrees Celcius. Calculate delta-H (in kJ/mol NaOH) for the solution process "NaOH yields Na(+)(aq) + OH(-)(aq)". Assume that the specific heat of solution is thesame as that of pure water."</p>

Answer 1

The answer is ΔH = -44.4 kJ/mol NaOH. See solution below:

Mass of solution = mass of NaOH + mass of water

= 6.50 + 100.0 = 106.5 g

Specific heat capacity of solution = specific heat capacity of water = 4.184 J/g^oC

Heat released by reaction = heat absorbed by solution

= mass x specific heat capacity x temperature change of solution

= 106.5 x 4.184 x (37.8 - 21.6)

= 7219 J

Moles of NaOH = mass/molar mass of NaOH

= 6.50/40.00 = 0.1625 mol

ΔH = -heat released by reaction/moles of NaOH

= -7219/0.1625

= -4.44 x 10⁴ J/mol = -44.4 kJ/mol

Note that ΔH is negative as the reaction is exothermic and heat is released