Answer:
Increase the pressure of the gas
Explanation:
According to the Pressure law, for a fixed mass of gas, at a constant volume (V), the pressure (P) is directly proportional to the absolute temperature (T).
From the kinetic molecular theory, gases are composed of particles which are in constant motion, colliding with themselves as well as with the walls of their container.
When the temperature of these gas molecules is increased, the molecules acquire more kinetic energy and the rate of collisions increases. Since the container cannot expand, the increase in pressure is due to the increase in collisions between the molecules of the gas as well as with the walls of their container.
Molar mass Pb = 207.2 g/mol
1 mole Pb ------------- 207.2
? mole Pb ------------ 9.51 x 10³
moles = 9.51 x 10³ * 1 / 207.2
moles = 9.51 x 10³ / 207.2
= 45.89 moles
hope this helps!
A positive cahnge of enthalpy, ΔH rxn = + 55 kJ/mol, for the forward reaction means that the reaction is endothermic, i.e. the reactants absorb energy and the products are higher in energy.
Activation energy is the difference in the energy level of the reactants and the peak in the potential energy diagram (the energy of the transition state).
For an endothermic reaction, the products will be closer in energy to the transition state than what the reactans will be; so, the activation energy of the reversed reaction is lower than the activation energy of the forward reaction.
Activation energy of reverse and forward reactions is related by:
Activation energy of reverse rxn = Activation energy of forward rxn - ΔH rxn
=> Activiation energy of reverse rxn = 102 kJ/mol - 55 kJ/mol = 47 kJ/mol
Answer: 47 kJ/mol
(a) One form of the Clausius-Clapeyron equation is
ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:
Solving for ΔHv:
- ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
- ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:
- ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
- 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
- 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.
