Answer:
Part 1: - 1.091 x 10⁴ J/mol.
Part 2: - 1.137 x 10⁴ J/mol.
Explanation:
Part 1: At standard conditions:
At standard conditions Kp= 81.9.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(81.9)) = - 1.091 x 10⁴ J/mol.
Part 2: PICl = 2.63 atm; PI₂ = 0.324 atm; PCl₂ = 0.217 atm.
For the reaction:
I₂(g) + Cl₂(g) ⇌ 2ICl(g).
Kp = (PICl)²/(PI₂)(PCl₂) = (2.63 atm)²/(0.324 atm)(0.217 atm) = 98.38.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(98.38)) = - 1.137 x 10⁴ J/mol.
With every electron stationed in its own orbital or paired off with each other in the higher energy level, the energy level is balanced and stable. The atoms that utilize this exception are Molybdenum, Chromium, Gold, Silver, and Copper.
Answer :
Charles's Law : It is defined as the volume is directly proportional to the temperature of the gas at constant pressure and number of moles.
Mathematically,
Boiling water bath Cool bath 1 Cool bath 2
Temperature (⁰C) 99 17 2
Temperature (K)(T) 273+99=372 273+17=290 273+2=275
Volume of water 0.0 27.0 34.0
in cool flask (mL)
Volume of water= 135.8 135.8 135.8
Air in flask (mL)
Volume of air 135.8 108.8 101.8
in cool flask (V)
The graph volume versus temperature for a gas is shown below.
Answer:
B. water displacement is used "graduated cylinder is filled with water (100 mL) and the object is then put inside. ... If the new water level is (120 mL) we now know that the object has a volume of 20 mL."
Explanation:
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
