Answer and Explanation:
- For the following reaction: H2(g) + F2(g) ⇌ 2HF(g) the values of K and Kp are not the same ⇒ FALSE
Step-by-step explanation: the relation between Kc and Kp is given by the following expresion:
Kp= Kc (RT)ⁿ
Where R is the gas constant, T is the temperature and n (also Δn) is the mol change. In this case, the mol change is 0, thus Kc is the same as Kp:
Δn= product mol in gas phase- reactants mol in gas phase
Δn= 2 mol - (1 mol + 1 mol)= 2 mol - 2 mol = 0
Kp= Kc(RT)⁰= Kc (1)= Kc
- The value of K at constant temperature does not depend on the amounts of reactants and products that are mixed together initially ⇒ TRUE
Step-by-step explanation: The value of the equilibrium constant depend on the temperature. When initial amounts of reactants and productes are mixed, whatever they are, they will evolutionate to reach the equilibrium and only the concentrations in the equilibrium determine the value of the constant (it is a fixed value).
- For a reaction with K >> 1, the rate of the forward reaction is less than the rate of the reverse reaction at equilibrium ⇒ FALSE
Step-by-step explanation: at equilibrium, the rate of the forward reaction and the rate of the reverse reaction are equal. If a reation have a K >> 1, that means that at equilibrium, there are significantly more amount of products than reactants (the reaction lies far to to the right)
- For the following reaction: CaCO3(s) ⇌ CaO(s) + CO2(g) the [CaCO3] should not appear in the equilibrium expression⇒ TRUE
Step-by-step explanation: In the equilibrium constant, pure solids and liquids are not included, because it is considered that their amounts do not affect the equilibrium (their activities are equal to zero). In this case, both CaCO₃ and CaO are solids, so their amounts are not included in the equilibrium constant expression.
- A large value of K means the equilibrium position lies far to the left⇒ FALSE
Step-by-step explanation: a large value of K means that there is more amount of products than reactants in the equilibrium, thus the equilibrium lies far to the right (the side of the products).