Answer to Question #169285 in General Chemistry for Dawson

Question

When 28.2 mL of 0.500 M H₂SO₄ is added to 28.2 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.). Do not answer in scientific notation.

Answer 1

$H_2SO_4 + 2KOH \rightarrow K_2SO_4 + H_2O$

Concentration of $H_2SO_4$ = 0.500M

Volume of $H_2SO_4$ = 0.028L

No. of moles of $H_2SO_4$ = $Concentration\times Volume$

= $0.028\times 0.500$

= $0.014$ mol

Concentration of KOH = 1M

Volume of KOH = 0.028L

Hence, number of moles of KOH = $0.028\times 1$

= $0.028$

The mole ratio is $1:2$

So all the quantity is used both for $KOH$ and $H_2SO_4$

$Mass = Density\times Volume$

$V_{total} = 56.4mL$

d = 1 g/mL

Hence, m = 56.4 g

$c= 4.2$ J/gK

$\Delta T$ = 30.17 - 23.5

$\Delta T = 6.67^ \circ C$

$Q = c\times \Delta T \times m$

= $4.2\times 6.67\times 56.4$

= $1579.98$ J

$\Delta H = \dfrac{Q}{n}$

= $\dfrac{1579.98}{0.04}$

= $39499.74$ J