Answer: Option (4) is the correct answer.
Explanation:
It is known that equilibrium constant is represented as follows for any general reaction.
K =
As equilibrium constant is directly proportional to the concentration of products so more is the value of equilibrium constant more will be the number of products formed.
As a result, more is the time taken by the reaction to reach towards equilibrium. Whereas smaller is the value of equilibrium constant more rapidly it will reach towards the equilibrium.
Thus, we can conclude that cases where K is a very small number will require the LEAST time to arrive at equilibrium.
Answer:
3.91 × 10⁴ J/mol
Explanation:
We can calculate the heat of solution using the following expression.
Q = c × m × ΔT
where,
c is the specific heat capacity of the solution
m is the mass of the solution
ΔT is the change in the temperature
The mass of LiCl is:
The mass of the solution is:
m = mLiCl + mH₂O = 10.6 g + 200.0 g = 210.6 g
Q = c × m × ΔT = (4.184 J g⁻¹ °C⁻¹) × 210.6 g × 11.08 °C = 9763 J
In a constant pressure calorimeter, the molar enthalpy of solution for LiCl is:
Considering the Hess's Law, the enthalpy change for the reaction is -84.4 kJ.
<h3>Hess's Law</h3>Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.
<h3>Enthalpy change for the reaction in this case</h3>In this case you want to calculate the enthalpy change of:
2 C (graphite) + 3 H₂(g) → C₂H₆(g)
which occurs in three stages.
You know the following reactions, with their corresponding enthalpies:
Equation 1: C₂H₆(g) + O₂(g) → 2 CO₂(g) + 3 H₂O(l) ; ΔH° = –1560 kJ
Equation 2: H₂(g) + O₂(g) → H₂O(l) ; ΔH° = –285.8 kJ
Equation 3: C(graphite) + O₂(g) → CO₂(g) ; ΔH° = –393.5 kJ
Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.
In this case, first, to obtain the enthalpy of the desired chemical reaction you need 2 moles of C(graphite) on reactant side and it is present in third equation. In this case it is necessary to multiply it by 2 to obtain the necessary amount. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 2, the variation of enthalpy also.
Now, you need 3 moles of H₂(g) on reactant side and it is present in second equation. In this case it is necessary to multiply it by 3 to obtain the necessary amount and the variation of enthalpy also is multiplied by 3.
Finally, 1 mole of C₂H₆(g) must be a product and is present in the first equation. Since this equation has 1 mole of C₂H₆(g) on the reactant side, it is necessary to locate the C₂H₆(g) on the reactant side (invert it). When an equation is inverted, the sign of delta H also changes.
In summary, you know that three equations with their corresponding enthalpies are:
Equation 1: 2 CO₂(g) + 3 H₂O(l) → C₂H₆(g) + O₂(g); ΔH° = 1560 kJ
Equation 2: 3 H₂(g) + O₂(g) → 3 H₂O(l) ; ΔH° = –857.4 kJ
Equation 3: 2 C(graphite) + 2 O₂(g) → 2 CO₂(g) ; ΔH° = –787 kJ
Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:
2 C (graphite) + 3 H₂(g) → C₂H₆(g) ΔH= -84.4 kJ
Finally, the enthalpy change for the reaction is -84.4 kJ.
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