The gas phase decomposition of dimethyl ether follows first order kinetics:
CH₃ – O CH₃ (g) → CH₄ (g) + H₂(g) + CO(g)
The reaction is carried out in a constant volume container at 500°C and has a half life of 14.5 minutes. Initially only dimethyl ether is present at a present of 0.40 atmosphere. What is the total pressure of the system after 12 minutes?
(Assume ideal gas behaviour).

Solution

The gas phase decomposition of dimethyl ether follows first order kinetics:
CH₃ – O CH₃ (g) → CH₄ (g) + H₂(g) + CO(g)

we have given
Volume is constant
temperature is 500⁰ C
half life is =14.5 min

PV = nRT

\frac{n}{V} = \frac{P}{RT}

a = \frac{P}{RT} = \frac{0.4}{0.082 \times 773} = 6.31 \times 10^{- 3} mol L^{- 1} K = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{14.5} = 4.78 \times 10^{- 2} \min^{- 1} K = \frac{2.303}{12} \log \frac{a}{a - x}

4.78 \times 10^{- 2} = \frac{2.303}{12} \log \frac{a}{a - x} \frac{a}{a - x} = 1.77446 a - x = \frac{6.31 \times 10^{- 3}}{1.77446} moles L^{- 1} = 3.556 \times 10^{- 3} moles L^{- 1} ∴ x = (6.310 - 3.556) \times 10^{- 3} moles / L = 2.754 \times 10^{- 3} moles / L ∴ After 12 minutes .

Total number of moles

L^{- 1}

= a + 2 x = 6.31 \times 10^{- 3} + 2 \times 2.754 \times 10^{- 3} = 11.818 \times 10^{- 3}

P = \frac{n}{V} RT = 11.818 \times 10^{- 3} \times 0.082 \times 773 = 0.749 atm .

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Chemical Kinetics

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The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

What will be the effect of temperature on rate constant?

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