Question

What is the molar concentration of H atoms at equilibrium if the equilibrium concentration of H2 is 0.28 M? Express your answer to two significant figures and include the appropriate units.

Answer 1

Each H2 molecule dissociates into two H atoms, so the molar concentration of H atoms at equilibrium is twice the concentration of H2, which calculates to 0.56 M.

Step-by-step explanation:

The molar concentration of hydrogen atoms (H atoms) at equilibrium when the equilibrium concentration of hydrogen gas (H2) is 0.28 M can be determined by understanding that each molecule of H2 dissociates into two H atoms. Since H2 exists as a diatomic molecule, the molar concentration of H atoms would be twice the molar concentration of H2 molecules at equilibrium. The calculation is as follows:

[H atoms] = 2 × [H2]

Therefore, the molar concentration of H atoms at equilibrium is:

[H atoms] = 2 × 0.28 M = 0.56 M

This value has two significant figures, as requested

Answer 2

0.56M

Step-by-step explanation:

Molar concentration is defined as the ratio between moles of solute and volume in liters of solution.

In a 0.28M H₂ there are 0.28moles of H₂ per liter of solution.

Now, in 1 molecule of H₂ there are 2 atoms of H. Following this idea, in 0.28 moles of H₂ there are 0.28*2 = 0.56 moles of H atoms.

Thus, molar concentration of H atoms in a 0.28M H₂ is 0.56M