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)

Question

Jair Sweeney

Chemistry

16 January, 17:34

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)

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Draven Patel · Answer 1

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