Answer:
The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.
Explanation:
..[1]
..[2]
..[3]
The enthalpy of reaction for the reaction of chlorine with ozone can be calculated by using Hess's law:
[2] - [1] = [3]
The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.
A curve of temperature vs. time for the entire heating process.
The sample is heated up to 100.°C, therefore, the heat and time required to heat the sample to its boiling point, the heat and time required to boil the sample, and the heat and time required to heat the sample from its boiling point to 100.°C are needs to be calculated.
i ) Calculating the heat and time required to heat the sample to its boiling point:
Boiling point = 85°C
C(liquid) = 2.5 J/g °C
The heat required up to melting the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from -20°C to 85°C can be calculated as,
Therefore, T f = 85°C and T i = - 20°C
Plug in the values in the specific heat formula to calculate the heat energy required to heat the sample to its melting point,
q3 = 25 g × 2.5 J/g °C × [85 - (-20)]°C
= 25 J/°C ×[85+20]°C
= 6562.5 J
The total heat energy required for heating the sample from initial temperature to boiling point is:-
q1 + q2 + q3 = 500 J + 4500 J + 6562.5 J
= 11562.5 J
The Rate of heating = 450 J/min
450. J = 1 min
11562.5 J = ? min
11562.5 J × 1min/450 J = 25.69 min
ii) Calculating the heat and time required to boil the sample:
∆H Vap = 500 J/g
The boiling is the phase change from liquid to gas at 85°C, therefore, the heat required to boil the sample can be determined
q4= m × ∆Hvap
= 25 g × 500 J/g
= 12500 J
Thus, total heat required to this phase change is q1 + q2 + q3 + q4 = 500 J + 4500 J +6562.5 J + 12500 J = 24062.5 J
The Rate of heating = 450 J / min
450 J = 1 min
24062.5 J = ? min
24062.5J × 1min / 450 J = 53.47 min
iii) Calculating the heat and time required to heat the sample from its boiling point to 100°C
C gas = 0.5 J / g °C
The heat required to boil the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from 85°C to 100°C can be calculated as,
Therefore, T f = 100.°C and T i = 85°C
q5 = 25 g × 0.5 J / g °C × [100 - 85] °C
= 25 J / °C ×15 °C
= 187.5 J
The total heat energy required for heating the sample from initial temperature to 100°C is
q1 + q2 + q3 + q4 + q5 = 500 J + 4500 J + 2625J + 12500 J + 187.5 J
=24250 J
The Rate of heating = 450 J / min
450. J = 1 min
24250 J=? min
Thus, heating the sample to 100.°C takes a total of 53.89 min.
iv) Draw a curve of temperature vs. time for the entire heating process:-
Temperature °C Temperature K Heat energy (J) Time (min)
-40 °C 233 0 0
-20 °C 253 500 1.11
Melting -20 °C 253 5000 11.11
85 °C 358 11562.5 25.69
Boiling 85 °C 358 24062.5 53.475
100 °C 373 24250 53.89
Hence, the graph for the result is in the image.
Learn more about temperature here:-brainly.com/question/24746268
#SPJ4
Answer:
a) ΔGrxn = 6.7 kJ/mol
b) K = 0.066
c) PO2 = 0.16 atm
Explanation:
a) The reaction is:
M₂O₃ = 2M + 3/2O₂
The expression for Gibbs energy is:
ΔGrxn = ∑Gproducts - ∑Greactants
Where
M₂O₃ = -6.7 kJ/mol
M = 0
O₂ = 0
b) To calculate the constant we have the following expression:
Where
ΔGrxn = 6.7 kJ/mol = 6700 J/mol
T = 298 K
R = 8.314 J/mol K
c) The equilibrium pressure of O₂ over M is:
Answer:
Explanation:
mole of O₂ =
= .25 moles
mole of CO₂
=
= .1818 moles
moles of SO₂
= .125 moles
Total moles of gas
= .5568 moles.
total volume of gas mixture
= 22.4 x .5568 liter ( volume of one mole of any gas = 22.4 liter)
= 12.47 liter.
gas will exert partial pressure according to their mole fraction
gas having greatest no of moles in the total mole will have greatest mole fraction so
O₂ will have greatest partial pressure.