Question

At \(800^{\circ} \mathrm{C}, K=2.2 \times 10^{-4}\) for the following reaction $$ 2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) $$ Calculate \(K\) at \(800^{\circ} \mathrm{C}\) for (a) the synthesis of one mole of \(\mathrm{H}_{2} \mathrm{~S}\) from \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) gases. (b) the decomposition of one mole of \(\mathrm{H}_{2} \mathrm{~S}\) gas.

Short answer

Answer: (a) K for the synthesis of one mole of H2S from H2 and S2 gases at 800°C is 4.69 × 10⁻². (b) K for the decomposition of one mole of H2S gas at 800°C is 4.69 × 10⁻².

Step-by-step solution

Rewrite the equation for (a)

For the synthesis of one mole of \(\mathrm{H}_{2} \mathrm{~S}\) from \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) gases, we can write the balanced equation as: $$ \mathrm{H}_{2}(g)+\frac{1}{2}\mathrm{S}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g) $$

Find the K value for (a)

To obtain K for (a), we must involve only half of the stoichiometry given in the original equation. The relationship between the reaction equilibrium constants can be established using the power rule as follows: \(\mathrm{K_{new}}=\mathrm{K^{1/n}}\) Here, the stoichiometry is reduced to half for the given balanced equation, which means \(n=2\): \(\mathrm{K_{new}} = \mathrm{K^{1/2}}\) Now, substitute the given K value and calculate K_new: $$ \mathrm{K_{new}}= (2.2 \times 10^{-4})^{1/2} $$ $$ \mathrm{K_{new}}= 4.69 \times 10^{-2} $$

Rewrite the equation for (b)

For the decomposition of one mole of \(\mathrm{H}_{2} \mathrm{~S}\) gas, we can write the balanced equation as: $$ \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \frac{1}{2}\mathrm{S}_{2}(g) + \mathrm{H}_{2}(g) $$

Find the K value for (b)

To obtain K for (b), we need to reduce two moles of \(\mathrm{H}_{2}\mathrm{~S}\) to one mole in the given balanced equation. Like in the previous case, we can establish the relationship between the equilibrium constants using the power rule: \(\mathrm{K_{new}}=\mathrm{K^{1/n}}\) Here, the stoichiometry of the given balanced equation is reduced to half, which means \(n=2\): \(\mathrm{K_{new}} = \mathrm{K^{1/2}}\) Now, substitute the given K value and calculate K_new: $$ \mathrm{K_{new}} = (2.2 \times 10^{-4})^{1/2} $$ $$ \mathrm{K_{new}} = 4.69 \times 10^{-2} $$ In conclusion: (a) K for the synthesis of one mole of \(\mathrm{H}_{2}\mathrm{~S}\) from \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) gases at 800°C is \(4.69 \times 10^{-2}\). (b) K for the decomposition of one mole of \(\mathrm{H}_{2}\mathrm{~S}\) gas at 800°C is \(4.69 \times 10^{-2}\).

Key concepts

Understanding Chemical Reaction Equilibrium

Chemical reaction equilibrium is a fundamental concept in chemistry that occurs when the rates of the forward and reverse reactions are equal, leading to a constant concentration of reactants and products over time. At this point, a system is said to be at equilibrium, and no further changes in the concentrations of reactants and products are observed.

Imagine a simple reaction where reactant A converts to product B. If initially there's a lot of A and little B, A will start turning into B, increasing its concentration. As B builds up, it starts turning back into A at a growing rate. Equilibrium is achieved when the rate at which A turns into B is the same as the rate at which B turns back into A, even though these reverse and forward processes continue to occur.

In practical terms, for students grappling with this concept, it's crucial to recognize that equilibrium does not imply that the reactants and products are present in equal amounts. It simply means that their amounts do not change over time because the two opposite reactions cancel each other out in terms of net effect.

Equilibrium Constant Calculation Breakdown

The equilibrium constant, denoted as K, is a numerical value that quantifies the position of a chemical reaction at equilibrium. It's a way to express the relative concentrations of the products to the reactants at this steady state, and it gives us a glimpse into the reaction's extent and direction.

The calculation of the equilibrium constant takes into account the balanced chemical equation. For a general reaction where reactants A and B form products C and D with stoichiometry aA + bB ⇌ cC + dD, the equilibrium constant K is given by:

\[ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
where the concentration of each species is raised to the power of its coefficient in the balanced equation. This mathematical expression is central to understanding where a reaction lies along the path from reactants to products.

When reaction conditions change (like a change in the stoichiometry), the equilibrium constant must be recalculated. As demonstrated in the textbook solution, modifying the equation's stoichiometry changes K according to the power rule, reflected in the new equilibrium constant \(K_{new} = K^{1/n}\). Students should note that this calculation assumes that the reaction occurs under the same temperature because K is temperature-dependent.

Reaction Stoichiometry and Equilibrium

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It tells us how much of each substance is involved in the reaction when it goes to completion. For students, it's crucial to understand that stoichiometry is the 'recipe' for chemical reactions; much like baking a cake requires certain proportions of ingredients, a reaction requires specific amounts of reactants to form products.

Consider the reaction given in the exercise:
\[2 \text{H}_2\text{S}(g) \rightleftharpoons 2\text{H}_2(g) + \text{S}_2(g)\]
Stoichiometric coefficients indicate that two moles of hydrogen sulfide decompose into two moles of hydrogen gas and one mole of sulfur gas under equilibrium conditions. During calculations related to equilibrium, like in the provided solution, it's essential to account for these stoichiometric ratios as they directly influence the equilibrium constant K.

When reaction stoichiometry changes, as in the construction or decomposition to form one mole of H2S instead of two, the equilibrium constant value must be adjusted. Understanding how to perform these adjustments allows students to determine the new equilibrium constants of the modified reactions.