Answer b I think so sorry if not is
18. <span>Answer is </span>
A<span>
<span>Since the enthalpy of reaction is positive, the
forward reaction is<span> an endothermic reaction which means the energy
is gained from the surrounding to happen the reaction. If the temperature
decreases, according to the </span></span>Le Chatelier's principle, the system tries to become equilibrium
by increasing temperature. Since forward reaction is endothermic (because of
the bond breaking), the backward reaction is exothermic (because of the bond
making) which releases the energy to the surroundings. This makes the increase
of temperature. So if the backward reaction is promoted because of the decrease
of temperature, then the concentration of H</span><span>₂ will decrease.</span>
<span>
</span>
19. Answer is A.
The reactant side
has 2 moles/molecules of reactants and the product side has 4 moles/molecules
of products which come from 1 N₂(g) and 3 H₂<span>(g). If the pressure is reduced in the system, according to the Le Chatelier's principle, the
system tries to increase the pressure. </span><span>Hence, forward
reaction is promoted because of the higher number
of molecules in product side. If the forward reaction is promoted, the
concentration of NH</span>₃(g) will decreased.
<span>20. </span>Answer is C.
If the concentration
of reactant is increased in the
system, according to the Le Chatelier's principle, the system tries
to reduce the concentration of that reactant. So if NH₃(g) concentration
is increased, then to be equilibrium, the forward reaction will be promoted.
Then the concentration of N₂<span>(g) will increase.</span>
<span> </span>
Answer:
- 10.555 kJ/mol.
Explanation:
∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.
Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).
∆H°rxn is the standard enthalpy change of the reaction (J/mol).
T is the temperature of the reaction (K).
∆S°rxn is the standard entorpy change of the reaction (J/mol.K).
∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants
<em>∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) </em>= (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.
∵ ∆S°rxn = ∑∆S°products - ∑∆S°reactants
<em>∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) </em>= (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) =<em> - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.</em>
<em></em>
∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.
<em>∴ ∆G°rxn = ∆H°rxn - T∆S°rxn </em>= (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = <em>- 10.555 kJ/mol.</em>
Atomic Number of Lithium is 3, so it has 3 electrons in its neutral state. Also, Li₂ will have 6 electrons. But the chemical formula we are given has a negative charge on it (i.e Li₂⁻) so there is an additional electron (RED) present on this compound. So, the total number of electrons are 7. The
MOT diagram for this compound is shown below. According to diagram we are having 4 electrons in Bonding Molecular Orbitals (
BMO) and 3 electrons in Anti-Bonding Molecular Orbitals (
ABMO). Bond Order is calculated as,
Bond Order = (# of e⁻s in BMO - # of e⁻s in ABMO) ÷ 2
Bond Order = (4 - 3) ÷ 2
Bond Order = 1 ÷ 2
Or,
Bond Order = 1/2Or,
Bond Order = 0.5
