Question

For the reaction PC15 (8) PC13 (g) + Cl2 (g) K = 0.0454 at 261 °C. If a vessel is filled with these gases such that the initial concentrations are [PC15) = 0.20 M, [PC13] = 0.20 M, and (Cl21 = 2.5 M, in which direction will a reaction occur and why? A) toward products because Qc = 0.56 B) toward reactants because Qc = 2.5 C) toward products because Qc = 2.8 D) toward reactants because Qc = 0.0454 E) it is at equilibrium because Qc = 1

Answer 1

Answer: The reaction proceed toward reactants because $Q_c$ is 2.5

Explanation:

$K_c$ is the constant of a certain reaction at equilibrium while $Q_c$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

The expression of $Q_c$ for above equation follows:

$Q_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

We are given:

$[PCl_3]=0.20M$

$[Cl_2]=2.5M$

$[PCl_5]=0.20M$

Putting values in above equation, we get:

$Q_c=\frac{0.20\times 2.5}{0.20}=2.5$

We are given:

$K_c=0.0454$

There are 3 conditions:

• When $K_{c}>Q_c$; the reaction is product favored.
• When $K_{c}$; the reaction is reactant favored.
• When $K_{c}=Q_c$; the reaction is in equilibrium.

As, $Q_c>K_c$, the reaction will be favoring reactant side.

Hence, the reaction proceed toward reactants because $Q_c$ is 2.5

Answer 2

Answer:

The correct answer is B.

The $K_{eq}$ is samller than $Q$ 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 $K_{eq}$

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:

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

The expression for $Q$ is written as:

$Q=\frac{[PCl_3][Cl_2]}{[[PCl_5]^1}$

$Q=\frac{0.20 M\times 2.5 M}{0.20 M}$

$Q=2.5$

Given : $K_{eq}$ = 0.0454

Thus as $K_{eq}$, the reaction will shift towards the left i.e. towards the reactant side.