Question

Which of the following is the expected product of the reaction of \(\mathrm{K}(s)\) and \(\mathrm{H}_{2}(g)\) ? (i) \(\mathrm{KH}(s)\), (ii) \(\mathrm{K}_{2} \mathrm{H}(s)\), (iii) \(\mathrm{KH}_{2}(s)\), (iv) \(\mathrm{K}_{2} \mathrm{H}_{2}(s)\), or (v) \(\mathrm{K}(s)\) and \(\mathrm{H}_{2}(g)\) will not react with one another.

Short answer

The expected product for the reaction between K(s) and H₂(g) is (i) KH (s).

Step-by-step solution

Potassium (K) is an alkali metal in Group 1 of the periodic table, so it has a valency of +1. Hydrogen (H) generally has a valency of +1 as well but can act as a reducing agent and accept an electron, making its valency -1 in certain situations. #Step 2: Analyze the possible products and their stoichiometry#

We have five options for the reaction's product: (i) KH(s): One K (+1) reacts with one H (-1), forming a 1:1 stoichiometry compound. (ii) K₂H(s): Two K (+1 each) react with one H (-1), forming a 2:1 stoichiometry compound. (iii) KH₂(s): One K (+1) reacts with two H (-1 each), forming a 1:2 stoichiometry compound. (iv) K₂H₂(s): Two K (+1 each) react with two H (-1 each), forming a 2:2 stoichiometry compound. (v) K(s) and H₂(g) will not react with one another: No reaction occurs between potassium and hydrogen. #Step 3: Determine the most likely product based on the valencies of the elements#

We know K has a valency of +1 and wants to lose one electron, while H will generally have a valency of +1 but can also have a valency of -1 and accept an electron. So it is likely that a compound will form in which K loses one electron and H gains one electron. This situation is achieved in product (i) KH(s), having a 1:1 stoichiometry and satisfying both elements' valency. Therefore, the expected product for the reaction between K(s) and H₂(g) is: #Final Answer#

(i) KH (s)

Key concepts

Stoichiometry

Stoichiometry is the process of determining the relative quantities of reactants and products in chemical reactions. It is essential for predicting the amounts of substances consumed and produced. In the given exercise, stoichiometry helps us understand that potassium (\(K\)) and hydrogen (\(H_2\)) can form a compound with a 1:1 ratio.

Here's why stoichiometry is important:

• It ensures the correct balance of atoms, satisfying the Law of Conservation of Mass.
• It helps predict the outcome of reactions, like forming \(KH\) from \(K\) and \(H_2\).
• It guides chemical synthesis and industrial production by determining how much reactant is needed.

In \(KH\), one potassium atom combines with one hydrogen atom, explaining the 1:1 stoichiometry observed in this reaction. Understanding these ratios allows chemists to predict products accurately.

Alkali Metals

Alkali metals belong to Group 1 of the periodic table and are known for their high reactivity. Potassium (\(K\)) is an alkali metal that reacts readily with other elements. These metals have unique properties:

• They have a single electron in their outer shell, which they tend to lose easily, forming cations with a +1 charge.
• They are soft and have low melting points compared to other metals.
• They react vigorously with water and other substances, which is why handling them requires care.

In the exercise, potassium's reactivity explains why it can readily react with hydrogen. Its tendency to lose an electron allows it to form stable ionic compounds like \(KH\), demonstrating typical alkali metal behavior.

Compound Formation

Compound formation involves the combination of two or more elements to create a new substance with different properties. It's essential to understand valency and electron transfer during this process.

In the reaction between potassium (\(K\)) and hydrogen (\(H_2\)), potassium donates an electron to hydrogen, transforming their individual properties into a stable compound, \(KH\).

• The electron transfer promotes stability, resulting in ionic bonding.
• Such reactions often release energy, indicating the formation of a strong bond.
• Ionic compounds generally have high melting and boiling points due to these strong interactions.

Understanding compound formation helps in predicting the feasibility and outcome of reactions, crucial for both academic studies and practical applications in industry.