A 100.0-mL sample of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M H_2SO_4 in a la

Question

A 100.0-mL sample of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M H_2SO_4 in a large Styrofoam coffee cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 23.7degree C. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the rmometer, the maximum temperature measured is 30.1 degree C. Assume that the density of the mixed solutions is 1.00 g/mL, that the specific heat of the mixed solutions is 4.18 J/(g degree C), and that no heat is lost to the surroundings. (a) Write a balanced chemical equation for the reaction that takes place in the Styrofoam cup. (b) Is any NaOH or H_2SO_4 left in the Styrofoam cup when the reaction is over? (c) Calculate the enthalpy change per mole of H_2SO_4 in the reaction.

Explanation / Answer

Answer – We are given, volume = 100 mL , [NaOH] = 1.00 M ,

Volume of H2SO4 = 50.0 mL , [H2SO4] = 1.00 M

Initial temp, ti = 23.7oC , tf = 30.1oC, density of solution = 1.00 g/mL , specific heat = 4.184 J/goC.

a)The balanced chemical reaction between NaOH and H2SO4

2 NaOH(aq) + H2SO4(aq)------> Na2SO4(aq) + 2 H2O(l)

b) Calculation of moles of both –

moles of NaOH = 1.00 M * 0.100 L = 0.100 moles

moles of H2SO4 = 1.00 M * 0.050 L = 0.050 moles

we know, from the above balanced reaction –

2 moles of NaOH needed 1 mole of H2SO4

So, 0.100 moles of NaOH needed 0.050 moles of H2SO4

So no any reactant left over in the Styrofoam cup

c) Calculation of mass of solution –

Total volume of solution = 100 + 50 = 150 mL

We know,

Density = mass / volume

So, mass = density x volume

               = 1.00 g/mL * 150 mL

               = 150 g

Now we need to calculate the absorbed by solution –

Heat, q = m *C*t

             = 150 g * 4.184 J/goC*(30.1oC – 23.7oC)

              = 4017 J

So change of enthalpy of reaction evolved = -4017

The enthalpy change per moles of H2SO4 = -4017 J / 0.050 moles

                                                                   = - 80333 J/mol

                                                                    = - 80.33 kJ/mol

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