Respuesta :
Answer:
∆H = 95.6 kJ/mol
Explanation:
Step 1: Data given
Volume of a 1.00 M NaOH = 100.2 mL
Volume a 1.00 M H2SO4 = 50.1 mL
The temperature of each solution before mixing is 22.45 °C
The maximum temperature measured is 30.90°C
the density of the mixed solutions is 1.00 g/mL
The specific heat of the mixed solutions is 4.18 J/g°C
Step 2: The balanced equation
2NaOH(aq) + H2SO4(aq) ==> Na2SO4(aq)+ 2H2O(l)
Step 3:
q = mC∆T
⇒q = the heat transfer = TO BE DETERMINED
⇒m = the mass of solution = 100.2 mL + 50.1 mL = 151.2 mL
151.3 ml * 1g/mL = 151.3 grams
⇒c is the specific heat of the solution = 4.18 J/g°C
⇒∆T = the change in temperature = T2 - T1 = 30.90 °C - 22.45 °C = 8.45 °C
q = 151.3 grams * 4.18 Jg°C * 7.65 °C
q = 4838.1 J
Step 4: Calculate ∆H per mole H2SO4
moles H2O4 = 0.0501 L*1.00 M = 0.0501 moles
∆H =4838.1 J / 0.0501 moles = 95569 J/mole = 95.6 kJ/mol
