Question

15.0 grams of NaOH was dissolved in 150.0 ml of 20.0 °C water. The temperature rose to 25.6 °C. Calculate the ΔH for the dissociation.

Answer 1

assuming density of water is 1 g/mL,

mass of water = density * volume

= 1 g/ml * 150.0 mL

= 150.0 g

Total mass of solution = 150.0 g + 15.0 g

= 165.0 g

Given:

m = 165 g

C = 4.184 J/g.oC

Ti = 20 oC

Tf = 25.6 oC

use:

Q = m*C*(Tf-Ti)

Q = 165.0*4.184*(25.6-20.0)

Q = 3866 J

Q = 3.866 KJ

This is heat given by solution.

So, sign of delta H would be negative

Molar mass of NaOH,

MM = 1*MM(Na) + 1*MM(O) + 1*MM(H)

= 1*22.99 + 1*16.0 + 1*1.008

= 39.998 g/mol

mass(NaOH)= 15.0 g

use:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(15 g)/(40 g/mol)

= 0.375 mol

delta H = -|q|/number of mol

= -(3.866 KJ)/0.375 mol

= -10.3 KJ/mol

Answer: -10.3 KJ/mol