6.03 Calorimetry Essay
Question # 1:
You are titrating 50 ml. of nitrous acid with sodium hydroxide (0.15 M). Equivalence is reached at 22.0 ml of NaOH delivered.
Calculate the pH of the solution in the flask:
a- before the beginning of the titration - after the delivery of 5 ml. of titrant b- at half-equivalence c-at equivalence d-after delivery of 23 ml. of titrant
Part A)
Reaction is:
HNO2 + OH- NO2- + H2O
Ka for Nitrous acid = 7.2E-4
Kw = Ka*Lb
Kb = 1.0E-14/7.2E-4
Kb = 1.38E-11
Moles of base:
(0.15)*(0.022) = 3.3E-3 moles
1:1 Ratio, So,
Moles of acid = 3.3E-3 moles
NaOH: (0.005)*(0.15) = 7.5E-4 moles
HNO2: 3.3E-3 – 7.5E-4 = 2.55 E-3 moles
Using Henderson- Hassel Balch equation:
Ph = 4.752 + log(7.5E-4/2.55E-3)
Ph =4.21
Part B)
Ph = 4.752 …show more content…
The change in entropy is situation to the heat transfer at constant pressure. Enthalpy is a state function. Usually enthalpy change calculated in relationships of internal energy change and pressure- volume correlation.
∆H= ∆E+ ∆(PV) Spontaneity of the reaction.
The standard for predicting spontaneity is founded on ΔG, the change in G, at constant temperature and pressure. If ΔG < 0, the process happens spontaneously. If ΔG > 0, the process is not spontaneous as written nonetheless occurs impulsively in the reverse direction. If ΔG = 0, the system is at equilibrium.
Explain how the temperature and the composition of the two phases (liquid and vapor) influence the spontaneity of the vaporization of water.
The energy mandatory for vaporization offsets the escalation in syndrome of the system. Therefore ΔG = 0, and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under normal situations. Both ΔH and TΔS are temperature reliant on, nonetheless the lines have conflicting slopes and cross at 373.15 K at 1 atm, where ΔH = TΔS. For the reason that ΔG = ΔH − TΔS, at this temperature ΔG = 0, representative that the liquid and vapor phases are in equilibrium. Question # 4:
Solution:
Formula is:
M1V1= M2V2
So,
V2 = 50*0.125/2
V2 = 3.125
You are titrating 50 ml. of nitrous acid with sodium hydroxide (0.15 M). Equivalence is reached at 22.0 ml of NaOH delivered.
Calculate the pH of the solution in the flask:
a- before the beginning of the titration - after the delivery of 5 ml. of titrant b- at half-equivalence c-at equivalence d-after delivery of 23 ml. of titrant
Part A)
Reaction is:
HNO2 + OH- NO2- + H2O
Ka for Nitrous acid = 7.2E-4
Kw = Ka*Lb
Kb = 1.0E-14/7.2E-4
Kb = 1.38E-11
Moles of base:
(0.15)*(0.022) = 3.3E-3 moles
1:1 Ratio, So,
Moles of acid = 3.3E-3 moles
NaOH: (0.005)*(0.15) = 7.5E-4 moles
HNO2: 3.3E-3 – 7.5E-4 = 2.55 E-3 moles
Using Henderson- Hassel Balch equation:
Ph = 4.752 + log(7.5E-4/2.55E-3)
Ph =4.21
Part B)
Ph = 4.752 …show more content…
The change in entropy is situation to the heat transfer at constant pressure. Enthalpy is a state function. Usually enthalpy change calculated in relationships of internal energy change and pressure- volume correlation.
∆H= ∆E+ ∆(PV) Spontaneity of the reaction.
The standard for predicting spontaneity is founded on ΔG, the change in G, at constant temperature and pressure. If ΔG < 0, the process happens spontaneously. If ΔG > 0, the process is not spontaneous as written nonetheless occurs impulsively in the reverse direction. If ΔG = 0, the system is at equilibrium.
Explain how the temperature and the composition of the two phases (liquid and vapor) influence the spontaneity of the vaporization of water.
The energy mandatory for vaporization offsets the escalation in syndrome of the system. Therefore ΔG = 0, and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under normal situations. Both ΔH and TΔS are temperature reliant on, nonetheless the lines have conflicting slopes and cross at 373.15 K at 1 atm, where ΔH = TΔS. For the reason that ΔG = ΔH − TΔS, at this temperature ΔG = 0, representative that the liquid and vapor phases are in equilibrium. Question # 4:
Solution:
Formula is:
M1V1= M2V2
So,
V2 = 50*0.125/2
V2 = 3.125