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atroni [7]
3 years ago
8

The isomerization of methylisonitrile to acetonitrile is first order in CH3NC. CH3NC(g) → CH3CN(g) The half life of the reaction

is 1.60 × 105 s at 444 K. What is the rate constant when the initial [CH3NC] is 0.030 M?
Chemistry
1 answer:
sasho [114]3 years ago
5 0

<u>Answer:</u> The rate constant for the given reaction is 4.33\times 10^{-6}s^{-1}

<u>Explanation:</u>

For the given chemical equation:

CH_3NC(g)\rightarrow CH_3CN(g)

We are given that the above equation is undergoing first order kinetics.

The equation used to calculate rate constant from given half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

The rate constant is independent of  the initial concentration for first order kinetics.

We are given:

t_{1/2} = half life of the reaction = 1.60\times 10^5s

Putting values in above equation, we get:

k=\frac{0.693}{1.60\times 10^5s}=4.33\times 10^{-6}s^{-1}

Hence, the rate constant for the given reaction is 4.33\times 10^{-6}s^{-1}

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<u>Answer:</u>

<u>For a:</u> The number of sodium ions in given amount of sodium sulfite are 2.65\times 10^{22}

<u>For b:</u> The number of sulfite ions in given amount of sodium sulfite are 1.325\times 10^{22}

<u>For c:</u> The mass of one formula unit of sodium ions is 2.09\times 10^{-22} grams

<u>Explanation:</u>

The chemical formula of sodium sulfite is  Na_2SO_3. It is formed by the combination of 2 sodium (Na^+) ions and 1 sulfite (SO_3^{2-}) ions

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of sodium sulfite = 2.80 g

Molar mass of sodium sulfite = 126 g/mol

Putting values in above equation, we get:

\text{Moles of sodium sulfite}=\frac{2.80g}{126g/mol}=0.022mol

  • <u>For a:</u>

Moles of sodium ions in sodium sulfite = (2 × 0.022) moles

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of particles

So, 0.022 moles of sodium sulfite will contain = (2\times 0.022\times 6.022\times 10^{23})=2.65\times 10^{22} number of sodium ions

Hence, the number of sodium ions in given amount of sodium sulfite are 2.65\times 10^{22}

  • <u>For b:</u>

Moles of sulfite ions in sodium sulfite = (1 × 0.022) moles

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of particles

So, 0.022 moles of sodium sulfite will contain = (1\times 0.022\times 6.022\times 10^{23})=1.325\times 10^{22} number of sulfite ions

Hence, the number of sulfite ions in given amount of sodium sulfite are 1.325\times 10^{22}

  • <u>For c:</u>

Molar mass of sodium sulfite = 126 g/mol

According to mole concept:

6.022\times 10^{23} number of formula units are present in 1 mole of a compound

Or, 6.022\times 10^{23} number of formula units of sodium sulfite have a mass of 126 grams

So, 1 formula unit of sodium sulfite will have a mass of = \frac{126}{6.022\times 10^{23}}\times 1=2.09\times 10^{-22}g

Hence, the mass of one formula unit of sodium ions is 2.09\times 10^{-22} grams

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Answer: 39.948 grams

Explanation:

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