Question

50 mL of .500 M NaOH is placed in a coffee-cup calorimeter at 25.00 degreees Celsius and 25 mL of .250 M H2SO4 is carefully added, also at 25.00 degrees Celsius. After stirring the final temperature is 27.21 degrees Celsius. Calculate q (soln) in J and the change in enthalpy, delta h in kj/mol of H2O formed. Assume that the total volume is the sum of the individual volumes, that d=1.00 g/mL, and that c= 4.184 J/g-k.

Answer 1

Given:

Volume of NaOH, V_NaOH = 50 ml

Molarity of NaOH, M_NaOH = 0.500 M

Volume of H₂SO₄, V_H2SO4 = 25 ml

Molarity of H₂SO₄, M_H2SO4 = 0.250 M

Initial temperature, T₁ = 25^oC

Final temperature, T₂ = 27.21^oC

specific heat, c = 4.184 J/g^oC

density, d = 1.00 g/ml

q = (specific heat) x (mass) x (change in temperature)

mass = density x total volume

total volume = 50 + 25 = 75 ml

q = 4.184 x (1 x 75)   x 2.21

q = 693.498 J

and q = 0.693 kJ

Balanced equation : NaOH + H2SO4 = H2O + Na2SO₄

Given moles of NaOH = M_NaOH x V_NaOH/1000 = 0.500 X 50/1000 = 0.025 mol

Given moles of H2SO4 = M_H2SO4 x V_H2SO4/1000 = 0.250 x 25/ 1000 = 0.00625 mol

But as per the balanced equation for the particular reaction, 1 mol of H₂SO₄ reacts with 2 moles of NaOH. we are provided with 0.00625 mol of H2SO4 which reacts with 0.00625 X 2 = 0.0125 mol of NaOH (H2SO4 is the limiting reagent).

since the reaction is exothermic, we need to assign negative charge