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3 years ago

At a certain temperature this reaction follows first-order kinetics with a rate constant of 2.01 s^âˆ’1

Chemistry

1 answer:

ddd [48]3 years ago

6 0

Answer:

0.80 seconds (2 significant figures)

Explanation:

The equation of the reaction is given as;

CICH2CH2Cl (g) --> CH2CHCI (g) + HCl(g)

Rate constant (k) = 2.01 s^-1

From the units of the rate constant, this is a first order reaction.

Initial Concentration = 1.34 M

t = ?

Final concentration = 20% of 1.34 = 0.268 M

The integrated rate law for a first order reaction is given as;

ln[A] = ln[A]o - kt

ln(0.268) = ln(1.34) - 2.01(t)

-2.01(t) = - 1.6094

t = 0.8007 ≈ 0.80 seconds (2 significant figures)

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3 years ago

How many moles of KNO3 are in 500 mL of 2. 0 M KNO3? mol KNO3.

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2 years ago

Butane (C4 H10(g), mc031-1.jpgHf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , mc031-2.jpgHf = –393.5 kJ/

ankoles [38]

The balanced chemical equation for the combustion of butane is:

$2C_{4}H_{10}(g) +13 O_{2}(g)-->8CO_{2}(g)+10H_{2}O(g)$

Δ $H_{reaction}^{0}$ = Σ $n_{products}$ Δ $H_{f}^{0}_{(products)}$ -Σ $n_{reactants}$ Δ $H_{f}^{0}_{(reactants)}$

= $[{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]$ =[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]

= -5315 kJ/mol

Calculating the enthalpy of combustion per mole of butane:

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Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol

Correct answer: -2657.5 kJ/mol

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3 years ago