At a particular temperature, \(K_{c}=1.6 \times 10^{-2}\) for $$ 2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) $$ Calculate \(K_{c}\) for each of the following reactions: (a) \(\frac{1}{2} \mathrm{~S}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)\) (b) \(5 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 5 \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{~S}_{2}(g)\)

Chemical equilibrium is a fundamental concept in chemistry. It occurs when a chemical reaction and its reverse reaction happen at the same rate. When equilibrium is reached, the concentrations of reactants and products remain constant over time. This doesn't mean the reactions stop; they continue, but no net change is observed.
A classic example is the reaction between hydrogen sulfide (H₂S), hydrogen (H₂), and sulfur (S₂) in the exercises provided. At equilibrium, the rate of H₂S breaking down to form H₂ and S₂ is exactly equal to the rate of H₂ and S₂ combining to form H₂S.
Understanding chemical equilibrium helps us predict how changing certain conditions (like pressure, temperature, or concentration) can affect the reaction's balance. This prediction is crucial not only for academic purposes but also in many industrial processes.

The equilibrium constant, denoted as K_c, gives us a quantifiable measure of a reaction's balance between products and reactants at equilibrium. It is derived from the ratio of the concentrations of the products to the reactants, each raised to the power of their coefficients in the balanced chemical equation.
For example, in the original exercise reaction:
\[2 \text{H}_{2}\text{~S}(g) \rightleftharpoons 2 \text{H}_{2}(g) + \text{S}_{2}(g) \]
the given equilibrium constant is K_c = 1.6 × 10^-2.
Knowing K_c allows chemists to determine how favorably a reaction proceeds toward products or reactants. If K_c is much greater than 1, the products are favored, while if K_c is much less than 1, the reactants are favored. Manipulating the original reaction (like reversing it or halving its coefficients) affects the equilibrium constant, as seen in the provided solutions:

• For reaction (a) (reversing and halving the reaction), the equilibrium constant becomes K_c'' = 7.9.
• For reaction (b) (multiplying the reaction by a factor of 5/2), the equilibrium constant is K_c'' = 1.05 × 10^-7.

Understanding how to manipulate a chemical reaction is crucial for calculating new equilibrium constants when the reaction conditions are changed. Below are key manipulations:

• Reversing a Reaction: When you reverse a chemical reaction, you take the reciprocal of the equilibrium constant. For example, if K_c for the forward reaction is 1.6 × 10^-2, for the reverse reaction it would be \(\frac{1}{1.6 \times 10^{-2}} = 62.5\).
• Changing the Reaction Coefficients: If you multiply all coefficients in the balanced equation by a common factor, you raise K_c to the power of that factor. This was exemplified in reaction (b), where multiplying by 5/2 led to the new equilibrium constant of \((1.6 \times 10^{-2})^{5/2} = 1.05 \times 10^{-7}\).

Manipulating the coefficients or direction of reactions helps predict and control chemical processes effectively. This concept is vital for both theoretical understanding and practical applications in chemistry.