Question

40.0 mL of 0.300 M NaOH are mixed with 40.0 mL of 0.300 M HBr in a calorimeter. The temperature increases from 25.00 °C to 27.00 °C. Assume that the specific heat of the solution is the same as that of pure water (4.18 J/(g•°C)) and that the density is the same as pure water (1.00 g/mL). Calculate ΔH per mole of reaction for the below chemical reaction. NaOH (aq) + HBr (aq) → NaBr (s) + H2O (l)

Answer 1

Answer:

Delta H per mole =55733.3J

Explanation:

volume of solution = 40ml +40ml =80ml

Mass of solution = 80g (since the density is the same as pure water (1.00 g/mL)

Temperature rise = (27-25)oc =2oc.

Heat liberated (Q) = -McT

= 80* 4.18* 2= -668.8J

Number of moles of NaOH in solution = 0.3x 40/1000 =0.012 mole.

Number of moles of HBr in solution =0.3x 40/1000 =0.012 mole.

When 0.012 mole of NaOH reacts with 0.012 mole of HBr, -668.8J was liberated.

Hence. 1 mole of NaOH reacts with 1mole HBr to liberate. =-668.8/0.012 =55733.3J