Question

when 21.7 ml of 0.500 m h2so4 is added to 21.7 ml of 1.00 m koh in a coffee-cup calorimeter ... Your question has been answered Let us know if you got a helpful answer. Rate this answer Question: When 21.7 mL of 0.500 M H2SO4 is added to 21.7 mL of 1.00 M KOH in a coffee-cup calorimeter at 23... When 21.7 mL of 0.500 M H2SO4 is added to 21.7 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH of this reaction. (Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water.) (d for water = 1.00 g/mL; c for water = 4.184 J/g·°C.) kJ/mol H2O ?

Answer 1

Explanation:

The given data is as follows.

  Molarity of $H_{2}SO_{4}$ = 0.500 M

    Volume $H_{2}SO_{4}$ solution = 21.7 mL

Hence, calculate the number of moles of sulfuric acid as follows.

No. of moles of $H_{2}SO_{4}$ = $molarity \times volume$

                              = $0.500 M \times 21.7 \times 10^{-3} L$

                              = 0.01085 mol

Now, calculate the moles of KOH as follows.

     No. of moles of KOH = $\text{molarity of KOH} \times \text{volume KOH in liter}$

                                     = $1.00 M \times 21.7 \times 10^{-3} L$

                                        = 0.0217 mol

As the reaction equation is as follows.

     $H_{2}SO_{4}(aq) + 2KOH(aq) \rightarrow K_{2}SO_{4}(aq) + 2H_{2}O (l)$

Calculate the number of moles of water as follows.          

     No. of moles of $H_{2}O$ formed = $2 \times moles of H_{2}SO_{4}$ = (moles of KOH)

                              = 0.0217 mol

Now, total volume of the solution is as follows.

Total volume of solution = (volume $H_{2}SO_{4}$) + (volume KOH)

                                          = (21.7 mL) + (21.7 mL)

                                         = 43.4 mL

Total mass of solution = $(\text{Total volume of solution}) \times (\text{density of solution})$

Total mass of solution = $43.4 mL \times 1.00 g/mL$

                                      = 43.4 g

Heat change of solution = $(\text{Total mass of solution}) \times (\text{specific heat of solution}) \times (\text{final temperature - initial temperature})$

                                       = $(43.4 g) \times (4.184 J/g^{o}C) \times (30.17^{o}C - 23.50^{o}C)$

                                       = 1211.18 J

As, heat change of reaction = -(Heat change of solution)

Therefore, heat change of reaction = -1211.18 J

Hence, calculate the change in enthalpy as follows.

           $\DeltaH = \frac{\text{Heat change of reaction}}{\text{moles of H_{2}O formed}}$

                                 = $\frac{-1211.18 J}{0.0217 mol}$

                                 = -55814.56 J/mol

or,                             = -55.8 kJ/mol            (as 1 kJ = 1000 J)

Thus, we can conclude that $\Delta H$ of the given reaction is  -55.8 kJ/mol.