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a_sh-v [17]
3 years ago
10

The rate constant, k, for a first-order reaction is equal to 4.3 × 10-4 s-1. What is the half-life for the reaction? The rate co

nstant, k, for a first-order reaction is equal to 4.3 × 10-4 s-1. What is the half-life for the reaction? 1.9 × 103 s 1.2 × 103 s 1.6 × 103 s 3.0 × 10-4 s
Chemistry
1 answer:
pantera1 [17]3 years ago
4 0

Answer:

1.6 × 10³ s

Explanation:

Let's consider the following generic reaction.

A → B

The rate law is:

rate=k \times [A]^{m}

where,

rate is the reaction rate

k is the rate constant

[A] is the molar concentration of the reactant A

m is the reaction order

When m = 1, we have a first-order reaction. We can calculate the half-life for this reaction using the following expression.

t_{1/2}=\frac{ln2}{k} =\frac{ln2}{4.3 \times 10^{-4}s^{-1} }=1.6 \times 10^{3}s

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<h3>Further explanation</h3>

Given

mass of sugar = 12.45 g

mass of water = 100 g

Required

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Mass of solution :

\tt mass~sugar+mass~water=12.45+100=112.45~g

Percent mass of sugar :

\tt \%mass=\dfrac{mass~sugar}{mass~solution}\times 100\%\\\\\%mass=\dfrac{12.45}{112.45}\times 100\%=11.07\%

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a sample of 3.00 g of so2 (g)originally in a 5.00 L vesselat 21 degee Celsius is transferred to a 10.0 L vessel at 26 degree Cel
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Answer:

1) The partial pressure of SO₂ gas in the larger container = 0.115 atm.

2) The partial pressure of N₂ gas in the larger container = 0.206 atm.

3) The total pressure in the vessel = 0.321 atm.

Explanation:

  • To calculate the partial pressure of each gas, we can use the general law of ideal gas: PV = nRT.

where, P is the partial pressure of the gas in atm,

V is the volume of the vessel in L,

n is the no. of moles of the gas,

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K.

<u><em>1) What is the partial pressure of SO₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (3.0 g)/(64.066 g/mol) = 0.047 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.047 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.115 atm.

<u><em>2) What is the partial pressure of N₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (2.35 g)/(28.0 g/mol) = 0.084 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.084 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.206 atm.

<u><em>3) What is the total pressure in the vessel?</em></u>

  • According to Dalton's law the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

<em>∵ The total pressure in the vessel = the partial pressure of SO₂ + the partial pressure of N₂.</em>

∴ The total pressure in the vessel = 0.115 + 0.206 = 0.321 atm.

5 0
3 years ago
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