Problem 17 - Chemistry: The Central Science

Question

(a) When the metallic element sodium combines with the nonmetallic element bromine, \(\operatorname{Br}_{2}(l)\), what is the chemical formula of the product? (b) Is the product a solid, liquid, or gas at room temperature? (c) In the balanced chemical equation for this reaction, what is the coefficient in front of the product?

Step-by-Step Solution

Answer

(a) The chemical formula of the product when sodium combines with bromine is NaBr. (b) The product, sodium bromide (NaBr), is a solid at room temperature. (c) In the balanced chemical equation, the coefficient in front of the product NaBr is 2: 2 Na + \(\mathrm{Br}_{2}(l)\) → 2 NaBr

Step 1: Writing the unbalanced chemical equation

Sodium (Na) is a metal, and it'll lose 1 electron to form a stable +1 cation, while bromine (Br) is a non-metal, and each bromine atom will gain 1 electron to form a stable -1 anion. When sodium combines with bromine, they form an ionic compound called sodium bromide (NaBr). The unbalanced equation for this reaction is: Na + \(\mathrm{Br}_{2}(l)\) → NaBr

Step 2: Balancing the chemical equation

To balance the chemical equation, we need to make sure the number of atoms of each element on the reactant side is equal to the number of atoms of the same element on the product side. In the unbalanced equation, we have: Reactants: 1 Na and 2 Br atoms. Products: 1 Na and 1 Br atom. To balance the bromine atoms, we need to change the coefficient of NaBr to 2: Na + \(\mathrm{Br}_{2}(l)\) → 2 NaBr Now, we also need to balance the sodium atoms by changing the coefficient of Na to 2: 2 Na + \(\mathrm{Br}_{2}(l)\) → 2 NaBr Now, the chemical equation is balanced.

Step 3: Determining the state of the product at room temperature

Sodium bromide (NaBr) is an ionic compound, and ionic compounds usually form solid crystalline structures at room temperature. So, the product is a solid at room temperature.

Step 4: Finding the coefficient in front of the product

In the balanced chemical equation, the coefficient in front of the product NaBr is 2: 2 Na + \(\mathrm{Br}_{2}(l)\) → 2 NaBr So, the answer is 2.

Key Concepts

Ionic Compounds

Ionic compounds are formed through the transfer of electrons from one atom to another, usually between metals and non-metals. This electron transfer results in the formation of positive and negative ions. Metals, like sodium (Na), tend to lose electrons, becoming positively charged ions or cations. On the other hand, non-metals, such as bromine (Br), gain electrons to form negatively charged ions or anions.

When sodium (Na) and bromine (\(\text{Br}_2\text{(}l\text{)}\)) react, they create an ionic bond. The resulting compound is sodium bromide (NaBr). In this process, sodium transfers its one electron to bromine, fulfilling the octet rule and stabilizing the atoms involved. Ionic bonds are strong due to the electrostatic attraction between the oppositely charged ions, leading to the formation of crystal lattice structures characteristic of ionic compounds.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry, ensuring that matter is conserved during a chemical reaction. It involves adjusting the coefficients, which are the numbers in front of chemical formulas, to make the number of each type of atom the same on both sides of the equation.

In the case of sodium reacting with bromine to form sodium bromide:

• Start with the unbalanced equation: \(\text{Na} + \text{Br}_2 \rightarrow \text{NaBr}\).
• Balance bromine atoms by putting a coefficient of 2 in front of NaBr: \(\text{Na} + \text{Br}_2 \rightarrow 2 \text{NaBr}\).
• Next, balance sodium by putting a 2 in front of Na: \(2 \text{Na} + \text{Br}_2 \rightarrow 2 \text{NaBr}\).

At this point, each element has an equal number of atoms on both sides of the reaction, and the equation is balanced, reflecting the conservation of mass.

Sodium Bromide

Sodium bromide (NaBr) is an ionic compound that results from the reaction between sodium and bromine. As a chemical compound, it comprises sodium cations (Na⁺) and bromide anions (Br⁻), bonded together in a crystal lattice. This structure is typical for ionic compounds, providing them with high melting and boiling points.

Sodium bromide is soluble in water, dissociating into its ions, which is a common behavior for many ionic compounds. This solubility is essential for numerous applications, from chemical synthesis to its use in medicinal formulations. Being a bulk commodity chemical, NaBr is used in various industries, such as pharmaceuticals and as a source of bromide ions in chemical reagents.

States of Matter at Room Temperature

The state of matter of a substance at room temperature is vital to understanding its physical properties. Most substances exist in one of three primary states: solid, liquid, or gas, under standard conditions, which are about 25°C (77°F) and 1 atm pressure.

Ionic compounds like sodium bromide (NaBr) typically exist as solids at room temperature. This is due to the strong ionic bonds and cohesive forces that hold the ions in a fixed, organized crystal lattice structure. This structure makes them brittle and hard and why they are generally non-conductive in solid form but conductive when dissolved in water.

On the contrary, elements like bromine are liquids at room temperature. Bromine is one of the few elements that is a liquid under these conditions, demonstrating the unique properties of different types of chemical compounds and elements.