Chemistry, 20.11.2019 05:31 tristanlindor5908

Amixture of 1.374g of h2 and 70.31g of br2 is heated in a 2.00 l vessel at 700 k. these substances react as follows: h2(g) + br2(g) < > 2hbr(g)
at equilibrium the vessel is found to contain 0.566g of h2.
a. calculate the equilibrium concentrations of h2, br2 and hbr.
b. calculate kc.

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Answer from: benchang531

a. 0.139 M → [H₂] ; 0.217 M → [Br₂] ; 0.01 M → [HBr]

b. Kc = 3.31x10⁻³

Explanation:

H2(g) + Br2(g) ⇄ 2HBr(g)

Initial 1.374 g 70.31 g -

reacts X X 2x

eq. (1.374 - x) (70.31-x) 2x

In equilibrium I see, the grams I initially had minus some mass which has reacted. In products I have the double of that mass, because the stoichiometry.

So I have the mass in equilibrium, of H2 and of course I can know the mass which has reacted.

1.374g - x = 0.556 g

1.374g - 0.556 g = x = 0.808 g (This is the mass which has reacted)

70.31g - 0.808 g = 69.502 g (Mass in equilibrium of Br2)

2 . 0.808 g = 1.616 g (Mass in equibrium of HBr)

By molar mass, we can kwow the moles.

Molar mass H2: 2 g/m

Moles = mass / molar mass → 0.556 g / 2 g/m = 0.278 moles

Molar mass Br2: 159.80 g/m

Moles = mass / molar mass → 69.502 g / 159.80 g/m = 0.434 moles

Molar mass HBr: 80.9 g/m

Moles = mass / molar mass → 1.616 g / 80.9 g/m = 0.02 moles

The moles are not molarity. In equilibrium, to calculate Kc we need molarity (moles/L). The moles we have calculated are in 2 L of mixture so:

moles / L = molarity

0.278 moles / 2L = 0.139 M → [H₂]

0.434 moles / 2L = 0.217 M → [Br₂]

0.02 moles / 2L = 0.01 M → [HBr]

Kc = [HBr]² / ([H₂] . [Br₂])

Kc = 0.01² / (0.139 . 0.217) = 3.31x10⁻³

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

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Amixture of 1.374g of h2 and 70.31g of br2 is heated in a 2.00 l vessel at 700 k. these substances r...

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