A mixture of 1.374 g of H_2 and 70.31 g of Br_2 is heated in a 2.00 L vessel at 700 K. These substances react as follows: H_2 (g) + Br_2 (g) 2 HBr (g) At equilibrium the vessel is found to contain 0.566 g of H_2. Calculate the equilibrium concentration of HBr. Express your answer using three significant figures. Use the following values in your calculations: molar mass of H_2 = 2.0159 g/mol and molar mass of Br_2 = 159.81 g/mol. Do not round any of your values in your calculations to the requested number of significant figures until the very last step.

Question

Annabelle Flores

Chemistry

7 February, 14:48

A mixture of 1.374 g of H_2 and 70.31 g of Br_2 is heated in a 2.00 L vessel at 700 K. These substances react as follows: H_2 (g) + Br_2 (g) 2 HBr (g) At equilibrium the vessel is found to contain 0.566 g of H_2. Calculate the equilibrium concentration of HBr. Express your answer using three significant figures. Use the following values in your calculations: molar mass of H_2 = 2.0159 g/mol and molar mass of Br_2 = 159.81 g/mol. Do not round any of your values in your calculations to the requested number of significant figures until the very last step.

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Alyson Terrell · Answer 1

0.400 M

Explanation:

The reaction given is

H₂ (g) + Br₂ (g) ⇄ 2HBr (g)

So, the stoichiometry is 1 mol of H₂ for 1 mol of Br₂ to form 2 moles of HBr.

By the molas masses given, the number of moles of each compound is (mass/molar mass):

nH₂ = 1.374/2.0159 = 0.681581427 mol

nBr₂ = 70.31/159.81 = 0.439959952 mol

The number of moles of H₂ in the equilibrium will be:

nH₂e = 0.566/2.0159 = 0.280767895 mol

So the number of moles that reacts is the initial less the equilibrium:

n = 0.400813531 mol

For the stoichiometry, the number of moles that is formed of HBr must be double, which will be the number of moles in equilibrium:

nHBr = 0.801627063

The molar concentration is the number of moles divided by the volume:

0.801627063/2.00

0.400 M