2995 Views
2926 Views
2941 Views
2909 Views
2314 Views
2341 Views
Acetaldehyde vapours when heated upto 791K it breaks into methane and carbon monoxide gases, the reaction appears to be of the first order but experimentally it is found that the reaction is of second order because two molecules of acetaldehyde take part in reaction so the reaction can be written as,$$CH_3CHO (g) \overset { 791K }{ \longrightarrow } CH_4 (g) + CO (g)$$As both reactants and products are in gaseous state, the progress of the reaction is followed by constantly checking pressure of the reaction chamber at different intervals of time.
Also as the one mole of reactant (acetaldehyde) produces two moles of products (methane and carbon monoxide). So the decrease in pressure of reactant will be equal to the two times increase in pressure of products.
So at the start only acetaldehyde is present in reaction chamber. Therefore the pressure at start \({ P }_{ 0 }\) is due to initial amount of acetaldehyde i.e. a. $$ \left\{ \text { Pressure at the start } \left( { P }_{ 0 } \right) \right\} \propto \left\{ \text { Initial concentration of acetaldehyde } \left( a \right) \right\} $$ $$ \therefore a \propto { P }_{ 0 } \qquad ... (1)$$ After time t, let the pressure of reactant (acetaldehyde) decreases by p. At the same time pressure of each product (methane and carbon monoxide) increase by p. So total increase in pressure is p + p = 2p$$ $$ So pressure of reaction chamber at time t is given by, $$ { P }_{ t } = { P }_{ C{ H }_{ 3 }CHO } + { P }_{ C{ H }_{ 4 } } + { P }_{ CO }$$ where \({ P }_{ C{ H }_{ 3 }CHO }\) is the pressure after time \(t = \left( { P }_{ 0 } - p \right) \) So, $${ P }_{ t } = \left( { P }_{ 0 } - p \right) + p + p { P }_{ t } = { P }_{ 0 } + p p = { P }_{ t } - { P }_{ 0 } \qquad ... (2)$$ Also at this time amount of reactant (acetaldehyde) decomposed (x) is equivalent to the decrease in pressure (p) at time t. $$ \left\{ \text { Decrease in pressure of acetaldehyde } \left( p \right) \right\} \propto \left\{ \text { Amount of acetaldehyde decomposed } \left( x \right) \right\} $$ $$ \therefore x \propto p$$ Let’s put the value of p from equation (2) $$ x \propto { P }_{ t } - { P }_{ 0 } \qquad ... (3)$$ Now we have both values of a and x from equation (1) and (3). Let’s put these values in rate constant equation for second order reaction, which is $$ k = \frac { 1 }{ t } \frac { x }{ a\left( a - x \right) } $$ $$ k = \frac { 1 }{ t } \frac { \left( { P }_{ t } - { P }_{ 0 } \right) }{ { P }_{ 0 }\left[ { P }_{ 0 } - \left( { P }_{ t } - { P }_{ 0 } \right) \right] } $$ $$ k = \frac { 1 }{ t } \frac { \left( { P }_{ t } - { P }_{ 0 } \right) }{ { P }_{ 0 }\left[ { P }_{ 0 } - { P }_{ t } + { P }_{ 0 } \right] } $$ $$ \boxed { k = \frac { 1 }{ t } \frac { \left( { P }_{ t } - { P }_{ 0 } \right) }{ { P }_{ 0 }\left( 2{ P }_{ 0 } - { P }_{ t } \right) } } $$ So from the initial pressure \(\left( { P }_{ 0 } \right) \) and pressure are time t i.e. \(\left( { P }_{ t } \right) \) we can calculate rate of reaction for thermal decomposition of acetaldehyde.
Numerical: The thermal decomposition of acetaldehyde was studied by Hinshelwood and Hutch is on at 518C. Starting with an initial pressure of 363mm of Hg, the following results were obtained at different intervals of time.
Time (sec): | 42 | 73 | 105 | 190 |
---|---|---|---|---|
Pressure, \({ P }_{ t }\) (mm): | 397 | 417 | 437 | 477 |
Solution: Given that, Initial pressure \({ P }_{ 0 } = 363 mm\) If the reaction is of second order, it must obey the equation. $$ \boxed { k = \frac { 1 }{ t } \frac { \left( { P }_{ t } - { P }_{ 0 } \right) }{ { P }_{ 0 }\left( 2{ P }_{ 0 } - { P }_{ t } \right) } } $$ 1) at time t = 42 sec Pressure, \({ P }_{ t } = 397\) mm $$ k = \frac { 1 }{ 42 } \times \frac { \left( 397 - 363 \right) }{ 363\left[ \left( 2\times 363 \right) - 397 \right] } = \frac { 1 }{ 42 } \times \frac { 34 }{ 363\left[ 726 - 397 \right] } $$ $$ k = \frac { 1 }{ 42 } \times \frac { 34 }{ 363 \times 329 } = \frac { 34 }{ 5015934 } = 6.78 \times { 10 }^{ -6 }$$ 2) at time t = 73 sec Pressure, \({ P }_{ t } = 417\) mm $$ k = \frac { 1 }{ 73 } \times \frac { \left( 417 - 363 \right) }{ 363\left[ \left( 2\times 363 \right) - 417 \right] } = \frac { 1 }{ 73 } \times \frac { 54 }{ 363\left[ 726 - 417 \right] } $$ $$ k = \frac { 1 }{ 73 } \times \frac { 54 }{ 363 \times 309 } = \frac { 54 }{ 8188191 } = 6.59 \times { 10 }^{ -6 }$$ 3) at time t = 105 sec Pressure, \({ P }_{ t } = 437\) mm $$ k = \frac { 1 }{ 105 } \times \frac { \left( 437 - 363 \right) }{ 363\left[ \left( 2\times 363 \right) - 437 \right] } = \frac { 1 }{ 105 } \times \frac { 74 }{ 363\left[ 726 - 437 \right] } $$ $$ k = \frac { 1 }{ 105 } \times \frac { 74 }{ 363 \times 289 } = \frac { 74 }{ 11015235 } = 6.72 \times { 10 }^{ -6 }$$ 4) at time t = 190 sec Pressure, \({ P }_{ t } = 477\) mm $$ k = \frac { 1 }{ 190 } \times \frac { \left( 477 - 363 \right) }{ 363\left[ \left( 2\times 363 \right) - 477 \right] } = \frac { 1 }{ 190 } \times \frac { 114 }{ 363\left[ 726 - 477 \right] } $$ $$ k = \frac { 1 }{ 190 } \times \frac { 114 }{ 363 \times 249 } = \frac { 114 }{ 17173530 } = 6.64 \times { 10 }^{ -6 }$$ As the value of k is almost constant, the reaction is of second order and the mean rate constant is \(6.68 \times { 10 }^{ -6 }\)
Numerical: In the thermal decomposition of acetaldehyde at 518°C the increase in pressure was determined at various times. The results were as follows, Calculate the velocity constant of the reaction assuming it to be second order.
Time (sec): | 0 | 420 | 730 | 1900 | 2420 |
---|---|---|---|---|---|
Pressure \({ P }_{ t } \) (mm): | 363 | 397 | 417 | 477 | 497 |
2820 Views
2817 Views
6717 Views
1401 Views
2239 Views
Apply to join the passionate instructors who share their expertise and knowledge with the world. You'll collaborate with some of the industry's best producers, directors, and editors so that your content is presented in the best possible light..