By Sunil Bhardwaj

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Decomposition of Nitrogen Pentoxide

Suppose Reading of the polarimeter at zero time \(= { r }_{ 0 }\) and reading of the polarimeter at any time t \(= { r }_{ t }\) $$ \therefore \text { Angle of rotation at time } t = \left( { r }_{ 0 } - { r }_{ t } \right) $$ The angle of rotation at any time will be dependent on the amount of product formed or amount of reactants reacted i.e. x. So we can write, $$ \left( { r }_{ 0 } - { r }_{ t } \right) \propto x \qquad ...(1) $$ Reading of the polarimeter at infinite time \(= { r }_{ \infty }\) $$ \therefore \text {Angle of rotation at infinite time } = \left( { r }_{ 0 } - { r }_{ \infty } \right) $$ At infinite time the amount of product formed will be maximum and the amount of product formed will depend on the amount of reactant taken initially, i.e. a. So we can say,$$ \left( { r }_{ 0 } - { r }_{ \infty } \right) \propto a \qquad ...(2)$$ So we have both the values of x and a lets substitute it in first order rate equation.$$ k = \frac { 2.303 }{ t } \log { \frac { a }{ a-x } } $$ $$ \therefore k = \frac { 2.303 }{ t } \log { \frac { \left( { r }_{ 0 } - { r }_{ \infty } \right) }{ \left[ \left( { r }_{ 0 } - { r }_{ \infty } \right) - \left( { r }_{ 0 } - { r }_{ t } \right) \right] } } $$ $$ k = \frac { 2.303 }{ t } \log { \frac { \left( { r }_{ 0 } - { r }_{ \infty } \right) }{ \left[ { r }_{ 0 } - { r }_{ \infty } - { r }_{ 0 } + { r }_{ t } \right] } } $$ $$ \boxed { k = \frac { 2.303 }{ t } \log { \frac { \left( { r }_{ 0 } - { r }_{ \infty } \right) }{ \left( { r }_{ t } - { r }_{ \infty } \right) } } } $$ This is the equation for the rate constant.

Numerical: The inversion of cane sugar was studied in 0.9N HCl at 25°C. The following polarimetric readings were obtained at different intervals of time:

Time (min):07.181827.05\(\infty\)
Reading (deg):+24.09+21.41+17.74+15.00-10.74
Show that the inversion of cane sugar is a first order reaction.

Solution: If the reaction is of the first order, it must obey the equation, $$ k = \frac { 2.303 }{ t } \log { \frac { a }{ \left( a-x \right) } } $$ $$ \boxed { k = \frac { 2.303 }{ t } \log { \frac { \left( { r }_{ 0 } - { r }_{ \infty } \right) }{ \left( { r }_{ t } - { r }_{ \infty } \right) } } } $$ In this case \({ r }_{ 0 } = +24.09\) and \({ r }_{ \infty } = -10.74\) $$ \therefore { r }_{ 0 } - { r }_{ \infty } = +24.09 - \left( -10.74 \right) = 34.83$$ 1) at time t = 7.18 min rotation \({ r }_{ t } = +21.41\) $$ k = \frac { 2.303 }{ 7.18 } \log { \frac { \left( 34.83 \right) }{ \left( 21.41 + 10.74 \right) } } $$ $$ = 0.321 \times \log { \frac { \left( 34.83 \right) }{ \left( 32.15 \right) } } $$ $$ = 0.321 \times \log { \left( 1.083 \right) } = 0.011{ min }^{ -1 }$$ 2) at time t = 18 min rotation \({ r }_{ t } = +17.74\) $$ k = \frac { 2.303 }{ 18 } \log { \frac { \left( 34.83 \right) }{ \left( 17.74 + 10.74 \right) } } $$ $$ = 0.128 \times \log { \frac { \left( 34.83 \right) }{ \left( 28.48 \right) } } $$ $$ = 0.128 \times \log { \left( 1.223 \right) } = 0.011{ min }^{ -1 }$$ 3) at time t = 27.05 min rotation \({ r }_{ t } = +15.00\) $$ k = \frac { 2.303 }{ 27.05 } \log { \frac { \left( 34.83 \right) }{ \left( 15.00 + 10.74 \right) } } $$ $$ = 0.085 \times \log { \frac { \left( 34.83 \right) }{ \left( 25.74 \right) } } $$ $$ = 0.085 \times \log { \left( 1.353 \right) } = 0.011{ min }^{ -1 }$$ Since the value of k is nearly constant, given reaction is of first order. The average value of the rate constant \(= 0.011{ min }^{ -1 }\)

Numerical:The inversion of cane sugar catalyzed by hydrochloric acid was followed in a polarimeter by determining the angle of rotation of the reaction mixture. The following results were obtained at 298K:

Time (min):0102030\(\infty\)
Reading (deg):45.543.5141.639.75-20
Show that the inversion of cane sugar is a first order reaction and calculate the specific reaction rate.

Ans: \(0.003{ min }^{ -1 }\)