3) Assume you mix 100.0 mL of 0.200 M NaOH solution with 50.0 mL of 0.400 M HBr

Question

3) Assume you mix 100.0 mL of 0.200 M NaOH solution with 50.0 mL of 0.400 M HBr solution in a coffee cup calorimeter. The temperature of both solutions before mixing was 22.50 degree C, and it rises to 24.28 degree C after the reaction is complete. (a) What type of enthalpy change would this reaction have? (Circle the correct answer) (i) enthalpy of formation (ii) enthalpy of neutralization (iii) enthalpy of acidification (iv) enthalpy of alkalinization (b) Calculate the enthalpy change of the reaction in kJ per mole of HBr. Assume the densities of the solutions are all 1.00 g/mL and the specific heat capacities of the final solution is 4.20 J/g.deg. (c) Based on the answer in (b) is this process endo- or exothermic?

Explanation / Answer

A) The Enthalpy change this reaction would have = Enthalpy of neutralization

First, write the balanced equation for the reaction:

HBr + NaOH --> NaBr + H2O

Next, convert to moles:

Moles NaOH= 100mL x (1L / 1000mL) x (0.200 M / 1L) = 0.0200 moles NaOH

Moles HBr = 50.0mL x (1L / 1000mL) x (0.400 M / L) = 0.0200 moles HBr

It is impossible to determine the limiting reactatant at this point, because both reactions use up the same amount of product (1:1 ratio). It can be either.

Assuming NaOH was limiting, when we react 0.0200 moles NaOH, we produce:

100.0 mL x (1.0g/mL) + 50.0mL x (1.0g/mL) = 150g of mixture

Therefore, the heat generated =

Q = m Cp dT

= 150g x (4.20 J/g

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