Respuesta :
Answer:
The correct answer is B.
The [tex]K_{eq}[/tex] is samller than [tex]Q[/tex] of the reaction . So,the reaction will shift towards the left i.e. towards the reactant side.
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_{eq}[/tex]
K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.
For the given chemical reaction:
[tex]PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)[/tex]
The expression for [tex]Q[/tex] is written as:
[tex]Q=\frac{[PCl_3][Cl_2]}{[[PCl_5]^1}[/tex]
[tex]Q=\frac{0.20 M\times 2.5 M}{0.20 M}[/tex]
[tex]Q=2.5[/tex]
Given : [tex]K_{eq}[/tex] = 0.0454
Thus as [tex]K_{eq}<Q[/tex], the reaction will shift towards the left i.e. towards the reactant side.
Answer: The reaction proceed toward reactants because [tex]Q_c[/tex] is 2.5
Explanation:
[tex]K_c[/tex] is the constant of a certain reaction at equilibrium while [tex]Q_c[/tex] is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.
For the given chemical reaction:
[tex]PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)[/tex]
The expression of [tex]Q_c[/tex] for above equation follows:
[tex]Q_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}[/tex]
We are given:
[tex][PCl_3]=0.20M[/tex]
[tex][Cl_2]=2.5M[/tex]
[tex][PCl_5]=0.20M[/tex]
Putting values in above equation, we get:
[tex]Q_c=\frac{0.20\times 2.5}{0.20}=2.5[/tex]
We are given:
[tex]K_c=0.0454[/tex]
There are 3 conditions:
- When [tex]K_{c}>Q_c[/tex]; the reaction is product favored.
- When [tex]K_{c}<Q_c[/tex]; the reaction is reactant favored.
- When [tex]K_{c}=Q_c[/tex]; the reaction is in equilibrium.
As, [tex]Q_c>K_c[/tex], the reaction will be favoring reactant side.
Hence, the reaction proceed toward reactants because [tex]Q_c[/tex] is 2.5
