Question

A 1M solution of glucose reaches dissociation equilibrium according to the equation $6HCHO\rightleftharpoons C_36H_{12}O_6$. What is the concentration of $HCHO$ at equilibrium? (Equilibrium constant = $6\times {{10}^{22}}$ )
(A) $1.6\times {{10}^{-8}}M$
(B) $3.2\times {{10}^{-6}}M$
(C) $3.2\times {{10}^{-4}}M$
(D) $1.6\times {{10}^{-4}}M$

Answer 1

Hint: An equilibrium constant (K) is the relationship between the concentration of reactants and products present at equilibrium in a reversible chemical process at a given temperature. It is represented as the ratio of the concentration of products and reactants.

Formula Used: For the reaction: $6A \rightleftharpoons B$
The equilibrium constant is $K=\frac{[B]}{{{[A]}^{2}}}$

Complete Step by Step Answer:
The given reaction is:
$6HCHO\rightleftharpoons C_6H_{12}O_6$
The dissociation of glucose to form formaldehyde at a very high value of equilibrium constant is very small.
Given that the equilibrium constant (K) is $6\times {{10}^{22}}$
And the concentration of glucose, $\left[ {{C}_{6}}{{H}_{12}}{{O}_{6}} \right]=1M$
At equilibrium,
The equilibrium constant is $K=\frac{\left[ {{C}_{6}}{{H}_{12}}{{O}_{6}} \right]}{{{\left[ HCHO \right]}^{6}}}$
                                                 $6\times {{10}^{22}}=\frac{1}{{{\left[ HCHO \right]}^{6}}}$
                                               ${{\left[ HCHO \right]}^{6}}=\frac{1}{6\times {{10}^{22}}}$
                                               $\left[ HCHO \right]={{\left[ \frac{1}{6\times {{10}^{22}}} \right]}^{\frac{1}{6}}}$
                                               $\left[ HCHO \right]=1.6\times {{10}^{-4}}M$
Thus, the concentration of HCHO at equilibrium is $1.6\times {{10}^{-4}}M$
Correct Option: (D) $1.6\times {{10}^{-4}}M$.

Additional Information: Chemical equilibrium is a state in which the rates of both the forward and backward reactions are equal and the concentration of both the reactants and products is constant. At equilibrium, there is no net change in the number of moles, although conversion from reactants to products or products to reactants is still occurring. According to the law of mass action, the value of the equilibrium constant, K_c, is constant at a constant temperature. The concentrations of reactants and products may vary, but the value for K_c remains the same.

Note: The value of the equilibrium constant always remains the same at constant temperature, although the concentration of reactants and products may change. In cases where the substances reacting are liquid, the equilibrium constant is calculated using their concentrations. Alternatively, if the reactants and products are gases, then the equilibrium constant is calculated using the partial pressure of each substance.