Question

Starting with 2 moles \[SO_2\] and 1 mole \[O_2\]​ in 1L flask, mixture required 0.4 mole \[MNO_4\]- in acidic medium.\[2SO_2 + O_2 \rightleftharpoons 2SO_3\] Hence, equilibrium constant $K_c$ is:
A. $80$
B. $2$
C. $!6$
D. $2.8$

Answer 1

Hint:Initially you must be aware of the conditions required for the equilibrium and what is equilibrium. Then it would be easy for you to calculate the above numerical by just putting the values in the formula mentioned below. Just take care of the error considering the reactant and product side.

Formula used:For the reaction $A+B \rightleftharpoons AB$
$K=\dfrac {[AB]} {[A][B]}$

Complete step-by-step solution:
The equilibrium constant of a chemical reaction (usually denoted by the symbol) provides insight into the relationship between the products and reactants when a chemical reaction reaches equilibrium.
Given:Volume of flask: 1L
Moles of:
\[MNO_4\]= 0.4
\[SO_2\]=2
\[O_2\]=1
0.4mole of \[MNO_4\]​=1 mole of \[SO_2\]​=1 mole of \[SO_3\]
​ \[2SO_2 + O_2 \rightleftharpoons 2SO_3\]
\[
  2\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;1\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;0 \\
  1\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;0.5\;{\text{ }}\;{\text{ }}\;\;1 \\
 \]
$K_c=\dfrac {[S{O_3}]^2}{{[S{O_2}]^2}[O_2]}$
$\Rightarrow K_c =12(0.5 \times 12)$
$\therefore K_c=2$

Note: Equilibrium is that the state during which market supply and demand balance one another, and as a result prices become stable. Generally, an over-supply of products or services causes prices to travel down, which ends up in higher demand—while an under-supply or shortage causes prices to travel up leading to less demand. The balancing effect of supply and demand leads to a state of equilibrium.Acids and bases have a chemical equilibrium in solution. At chemical equilibrium, the products and reactants have reached a state of balance. Reactions may still be taking place within the sample, but the forward and reverse reactions are taking place at the same rate, so the concentrations of the products and reactants are not changing with time.