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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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How many grams of solid NaOH are required to prepare a 400ml of a 5N solution? show your work!
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<u>Answer:</u> The mass of solid NaOH required is 80 g

<u>Explanation:</u>

Equivalent weight is calculated by dividing the molecular weight by n factor. The equation used is:

\text{Equivalent weight}=\frac{\text{Molecular weight}}{n}

where,

n = acidity for bases = 1 (For NaOH)

Molar mass of NaOH = 40 g/mol

Putting values in above equation, we get:

\text{Equivalent weight}=\frac{40g/mol}{1eq/mol}=40g/eq

Normality is defined as the umber of gram equivalents dissolved per liter of the solution.

Mathematically,

\text{Normality of solution}=\frac{\text{Number of gram equivalents} \times 1000}{\text{Volume of solution (in mL)}}

Or,

\text{Normality of solution}=\frac{\text{Given mass}\times 1000}{\text{Equivalent mass}\times \text{Volume of solution (in mL)}}         ......(1)

We are given:

Given mass of NaOH = ?

Equivalent mass of NaOH = 40 g/eq

Volume of solution = 400 mL

Normality of solution = 5 eq/L

Putting values in equation 1, we get:

5eq/L=\frac{\text{Mass of NaOH}\times 1000}{40g/eq\times 400mL}\\\\\text{Mass of NaOH}=80g

Hence, the mass of solid NaOH required is 80 g

4 0
3 years ago
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