Answer:
The equilibrium constant for the reversible reaction = 0.0164
Explanation:
At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.
The reaction is given as
A ⇌ B
Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]
The rate of forward reaction = |r₁| = k₁ [A]
The rate of backward reaction = |r₂| = k₂ [B]
(Taking only the magnitudes)
where k₁ and k₂ are the forward and backward rate constants respectively.
k₁ = 0.010 s⁻¹
k₂ = 0.0610 s⁻¹
|r₁| = 0.010 [A]
|r₂| = 0.016 [B]
At equilibrium, the rate of forward and backward reactions are equal
|r₁| = |r₂|
k₁ [A] = k₂ [B] (eqn 1)
Note that equilibrium constant, K, is given as
K = [B]/[A]
So, from eqn 1
k₁ [A] = k₂ [B]
[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164
K = [B]/[A] = (k₁/k₂) = 0.0164
Hope this Helps!!!
Answer:
ΔH° of the reaction is -747.54kJ
Explanation:
Based on gas law, it is possible to find the ΔH of a reaction using ΔH of half reactions.
Using the reactions:
<em>(1) </em>C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJ
<em>(2) </em>CO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJ
<em>(3) </em>H₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJ
<em>(4) </em>C(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJ
<em>(5) </em>CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ
The sum of 4×(4) + (5) gives:
4C(graphite) + 8H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)
ΔH° = -74.81 kJ
×4 - 890.3 kJ = -1189.54kJ
Now, this reaction - 4×(1) gives:
4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)
ΔH° = -1189.54kJ - 4×-110.5 = <em>-747.54kJ</em>
<em></em>
Thus <em>ΔH° of the reaction is -747.54kJ</em>
There must be an intramolecular force. The oxygen atoms are produced as a result of the breakdown of oxygen molecules. Intramolecular force is necessary to stop the oxygen (O2) in the air from changing into the O atom.
Which force causes attraction between O2 molecules?
The result is the London dispersion force, a fleeting attractive attraction, which is created when the electrons in two neighboring atoms occupy positions that temporarily cause the atoms to form dipoles. This interaction is commonly described by the phrase "induced dipole-induced dipole attraction".
What is the difference between intramolecular forces and intermolecular forces which type is stronger?
In general, intramolecular forces are greater than intermolecular forces. Ion-dipole interaction exerts the strongest intermolecular force, followed by hydrogen bonds, dipole-dipole interaction, and London dispersion. Examples. Hydrogen bonding forces, London dispersion forces, and dipole-dipole forces are the three different kinds of intermolecular interactions. The three different kinds of intramolecular forces are metal bonds, ionic bonds, and covalent bonds.
Learn more about intramolecular forces: brainly.com/question/28170469
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