3608 Views
3462 Views
3476 Views
3650 Views
3180 Views
3628 Views
A reaction is said to be of the first order if the rate of the reaction depends upon one concentration term only. E.g. $$ N_2O_2(g) \longrightarrow N_2O_4(g) + \frac12{ O }_{ 2 }(g) \text{ (First Order reaction) } $$
Suppose we start with a moles per litre of the reactant A $$ \underset { a \ moles/liter }{ A } \longrightarrow Products \qquad ... (at \ start)$$ After time t, suppose x moles per litre of it have reacted. Therefore the concentration of A remaining after time \(t = (a - x)\) moles per litre. $$ \underset { (a-x) \ moles/liter }{ A } \longrightarrow Products ... (after \ time \ t)$$ Then according to law of mass action, $$ \text{Rate of reaction } \propto \text{ Conc of A remaining after time t.} $$ $$ i.e., \frac { dx }{ dt } \propto (a-x)$$ or $$ \frac { dx }{ dt } = k(a-x) ... (1) $$ where k is called the Rate constant. $$ \therefore \frac { dx }{ (a-x) } = kdt$$ Integrating equation this equation, we get,$$ \int { \frac { dx }{ (a-x) } } = k\int { dt } $$ $$ -\ln { (a-x) } = kt + I \qquad ... (2)$$ where \(I\) is a constant of integration.
In the beginning, when t = 0, x = 0 (as no substance has reacted at the start of the reaction). Putting these values in equation (2), we get,$$ -\ln { (a-0) } = k\left( 0 \right) + I\\ -\ln { (a) } = I \qquad ... (3) $$ Substituting the value of I in equation (2), we get, $$ -\ln { (a-x) } = kt + \left[ -\ln { (a) } \right] $$ $$ -\ln { (a-x) } + \ln { (a) } = kt$$ $$ kt = \ln { (a) } - \ln { (a-x) } $$ $$ kt = \ln { \frac { a }{ a-x } } $$ $$ k = \frac { 1 }{ t } \ln { \frac { a }{ a-x } } $$ $$ \boxed { k = \frac { 2.303 }{ t } \log { \frac { a }{ a-x } } } \qquad ... (4)$$ This equation can also be written as, $$ \boxed { k = \frac { 2.303 }{ t } \log { \frac { { C }_{ 0 } }{ { C }_{ t } } } } ... (5)$$ Where \({ C }_{ 0 }\) is initial concentration and \({ C }_{ t }\) is Concentration at time t.
Unit of Rate Constant (k): Equation for the rate constant for first order reaction is, $$ \boxed { k = \frac { 2.303 }{ t } \log { \frac { a }{ a-x } } } $$ where \(\frac { a }{ a-x } \) is ratio of concentration thus no unit.$$ \therefore unit \ of \ k = \frac { 1 }{ time } = \frac { 1 }{ sec } = { sec }^{ -1 } \ or \ { min }^{ -1 }$$
Numerical: The catalytic decomosition of hydrogen peroxide in aqueous solution was followed at 298K by titrating aliquot volumes of the reaction solution against standard \(KMn{ O }_{ 4 }\) solution. The following results were obtained.
Time in min | 0 | 5 | 10 | 15 |
---|---|---|---|---|
Volm of KMnO_{4} (cm^{3}) | 37.5 | 33.5 | 29.9 | 26.7 |
Solution: Given that Volume of \(KMn{ O }_{ 4 } \) consumed will be equal to the conc of reactant remaining after that time. Thus the rate constant, $$ \boxed { k =\frac { 2.303 }{ t } \log { \frac { { C }_{ 0 } }{ { C }_{ t } } } } $$ when t = 5min, $$ k = \frac { 2.303 }{ 5 } \log { \frac { 37.5 }{ 33.5 } } =0.02256 { min }^{ -1 } $$ when t = 10min, $$ k = \frac { 2.303 }{ 10 } \log { \frac { 37.5 }{ 29.9 } } =0.02264 { min }^{ -1 } $$ when t = 15min, $$ k = \frac { 2.303 }{ 15 } \log { \frac { 37.5 }{ 26.7 } } =0.02264 { min }^{ -1 } $$ The average value of specific reaction rate is \(0.02261 { min }^{ -1 }\)
3036 Views
3460 Views
3499 Views
5361 Views
3175 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..