Question

When 50.0 mL of 0.500 M H2SO4 is added to 50.0 mL of 1.00 M KOH in a coffee cup calorimeter, the temperature of the solution rises from 25.10°C to 31.77°C. Calculate ΔH of this reaction (in kJ/mol KOH). 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 that for pure water (c = 4.184J/g°C).a. -112 kJ/mol KOHb. -2.79 kJ/mol KOHc. -74.9 kJ/mol KOHd. -1.86 kJ/mol KOHe. -55.8 kJ/mol KOH

Answer 1

The correct answer is option e.

Step-by-step explanation:

Total volume of the mixture of solution= 50.0 mL + 50.0 mL = 100.0 mL

Total mass of the solution = M

Density of the mixture of solution = density of pure water = d = 1 g/mL

`Mass =Density* Volume`

`m=1 g/mL* 100.0 mL = 100.0 g`

Specific heat of mixture of solution = Specific of water = c

c = 4.184 J/g°C

Change in temperature = ΔT = 31.77°C - 25.10°C =6.67°C

Heat gained by the mixture of solution = Q

`Q=mc\Delta T`

`Q=100.0 g* 4.184 J/g^oC* 6.67^oC=2,790.73 J`

`2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O`

Moles of KOH = n

Volume of KOH solution = 50.0 mL = 0.050 L

Molarity of the solution = 1.00 M

`n=1.00M* 0.050 L=0.050 mol`

The ΔH of this reaction:

`\Delta H=(-Q)/(n)=(-2,790.73 J)/(0.050 mol)=-55,814.56 J/mol=-55.8kJ/mol`

The ΔH of this reaction is -55.8 kJ/mol.