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Standard XII

Chemistry

Order of Reaction

At 518∘ C the...

Question

At $518^{\circ} C$ the rate of decomposition of a sample of gaseous acetaldehyde initially at a pressure of 363 Torr, was 1.00 Torr $s^{- 1}$ when 5% had reacted and 0.5 Torr $s^{- 1}$ when 33% had reacted. The order of the reaction is:

0.0

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Solution

The correct option is B 2
$\frac{r_{1}}{r_{2}} = [\frac{a - x_{1}}{a - x_{2}}]^{m}$
$\frac{1}{0.5} = [\frac{0.95}{0.77}]^{m}$
$2 = (1.41)^{m}$
$l o g 2 = m l o g (1.41)$
$\frac{l o g 2}{l o g 1.41} = m$
$m = \frac{0.3010}{0.149}$
~ 2

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Similar questions

Q. At

518^{o} C

, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of

363 T o r r

, was

1.00 T o r r s^{- 1}

when

5 %

had reacted and

0.5 T o r r s^{- 1}

when

33 %

had reacted. The order of the reaction is:

Q. For the reaction,

2 N O + H_{2} \to N_{2} O + H_{2} O

the value of -dp/dt is found to be

1.50 T o r r s^{- 1}

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0.25 T o r r s^{- 1}

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Q. The gaseous decomposition reaction:

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At 518∘C the rate of decomposition of a sample of gaseous acetaldehyde initially at a pressure of 363 Torr, was 1.00 Torr s−1 when 5% had reacted and 0.5 Torr s−1 when 33% had reacted. The order of the reaction is:

The correct option is B 2r1r2=[a−x1a−x2]m 10.5=[0.950.77]m 2=(1.41)m log2=m log(1.41) log2log1.41=m m=0.30100.149 ~ 2

At $518^{\circ} C$ the rate of decomposition of a sample of gaseous acetaldehyde initially at a pressure of 363 Torr, was 1.00 Torr $s^{- 1}$ when 5% had reacted and 0.5 Torr $s^{- 1}$ when 33% had reacted. The order of the reaction is:

The correct option is B 2
$\frac{r_{1}}{r_{2}} = [\frac{a - x_{1}}{a - x_{2}}]^{m}$
$\frac{1}{0.5} = [\frac{0.95}{0.77}]^{m}$
$2 = (1.41)^{m}$
$l o g 2 = m l o g (1.41)$
$\frac{l o g 2}{l o g 1.41} = m$
$m = \frac{0.3010}{0.149}$
~ 2