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SIZIF [17.4K]
3 years ago
6

Consider the reaction. c2h5oh(l)⟶ c2h6(g)+12o2(g). state what further data would be needed to obtain δrh∘ at 500∘c and 1 bar.

Chemistry
1 answer:
UkoKoshka [18]3 years ago
6 0
For this type of problem, it is essential for you to have a data on the standard heats of formation of the substances given. For elemental substances or diatomic gases, the standard heat of formation is 0. Standard means the temperature is at 0°C and pressure at 1 atm. Calculate the standard heat of reaction using:

ΔH°rxn = ∑(Stoichiometric coefficient×ΔHf of products) - ∑(Stoichiometric coefficient×ΔHf of reactants)
Then, use this equation to find the reaction at T = 500°C and P = 1 bar:

ΔHrxn = ΔH°rxn + [∑(Stoichiometric coefficient×Cp of products) - ∑(Stoichiometric coefficient×ΔHf of reactants)]ΔT
So, you also need the Cp or specific heat capacities of the substances.
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Consider the second-order reaction:
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Initial concentration of HI is 5 mol/L.

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

Explanation:

2HI(g)\rightarrow H_2(g)+I_2(g)


Rate Law: k[HI]^2


Rate constant of the reaction = k = 6.4\times 10^{-9} L/mol s

Order of the reaction = 2

Initial rate of reaction = R=1.6\times 10^{-7} Mol/L s

Initial concentration of HI =[A_o]

1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2

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Final concentration of HI after t = [A]

t = 4.53\times 10^{10} s

Integrated rate law for second order kinetics is given by:

\frac{1}{[A]}=kt+\frac{1}{[A_o]}

\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}

[A]=0.00345 mol/L

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

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