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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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6 0
4 years ago
27.4 g of Aluminum nitrite and 169.9 g of ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams
GarryVolchara [31]

<u>Answer:</u> The mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For aluminium nitrite:</u>

Given mass of aluminium nitrite = 27.4 g

Molar mass of aluminium nitrite = 41 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium nitrite}=\frac{27.4g}{41g/mol}=0.668mol

  • <u>For ammonium chloride:</u>

Given mass of ammonium chloride = 169.9 g

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonium chloride}=\frac{169.9g}{53.5g/mol}=3.176mol

The chemical equation for the reaction of aluminium nitrite and ammonium chloride follows:

Al(NO_2)_3+3NH_4Cl\rightarrow AlCl_3+3N_2+6H_2O

By Stoichiometry of the reaction:

1 mole of aluminium nitrite reacts with 3 moles of ammonium chloride

So, 0.668 moles of aluminium nitrite will react with = \frac{3}{1}\times 0.668=2.004mol of ammonium chloride.

As, given amount of ammonium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium nitrite is considered as a limiting reagent because it limits the formation of product.

Excess moles of ammonium chloride = (3.176 - 2.004) mol = 1.172 moles

Calculating the mass of ammonium chloride by using equation 1, we get:

Excess moles of ammonium chloride = 1.172 moles

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

1.172mol=\frac{\text{Mass of ammonium chloride}}{53.5g/mol}\\\\\text{Mass of ammonium chloride}=(1.172mol\times 53.5g/mol)=62.7g

Hence, the mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams

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