Amixture 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.

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

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

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