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Benzenediazonium chloride is an organic compound with the formula \((C_6H_5N_2)Cl\). It is a salt of a diazonium cation and chloride. It exists as a colourless solid that is soluble in polar solvents including water. It is the parent member of the aryldiazonium compounds, which are widely used in organic chemistry. Because the salt is unstable, it is not commercially available but is prepared upon demand.$$ (C_6H_5N_2)Cl + H_2O \longrightarrow C_6H_5OH + N_2 + HCl $$
It can be observed that, volume of \({ N }_{ 2 }\) collected at time t is depends on the amount of \((C_6H_5N_2)Cl\) decomposed.
[Amount of \((C_6H_5N_2)Cl\) decomposed (x)] \( \propto \) [Volume of \(N_2\) gas collected at any time (\(V_t\))]
i.e. $$ x \propto { V }_{ t } \qquad ...(1)$$ At infinite time the total \({ N }_{ 2 }\) gas collected will depend on the amount of \((C_6H_5N_2)Cl\) taken initially.
[Amount of \((C_6H_5N_2)Cl\) initially taken (a)] \( \propto \) [Volume of \({ N }_{ 2 }\) gas collected at infinite time \( { V }_{ \infty }\) ]
i.e. $$ a \propto { V }_{ \infty } \qquad ...(2)$$ Substituting these values in the first order equation. $$ k = \frac { 2.303 }{ t } \log { \frac { a }{ a-x } } $$ we get, $$ \boxed { k = \frac { 2.303 }{ t } \log { \frac { { V }_{ \infty } }{ { V }_{ \infty }-{ V }_{ t } } } } $$
Numerical: Benzene diazonium chloride in aqueous solution decomposes according to the equation $$ (C_6H_5N_2)Cl \longrightarrow C_{ 6 }H_{ 5 }Cl + N_{ 2 } $$ Starting with an initial concentration of 10g/litre, the volume of \(N_{ 2 }\) gas obtained at 50°C at different intervals of time was found to be as follows:
Time (minutes): | 6 | 12 | 18 | 24 | 30 | \(\infty\) |
---|---|---|---|---|---|---|
Vol. of \({ N }_{ 2 }\) evolved (ml): | 19.3 | 32.6 | 41.3 | 46.5 | 50.4 | 58.3 |
Solution: If the reaction is of the first order, it must obey the equation,$$ k = \frac { 2.303 }{ t } \log { \frac { a }{ a-x } } $$ or $$ \boxed { k = \frac { 2.303 }{ t } \log { \frac { { V }_{ \infty } }{ { V }_{ \infty }-{ V }_{ t } } } } $$ In the present case, $${ V }_{ \infty } = 58.3ml$$ The value of k at each time can be calculated as follows:
at time t = 6 min volume of \({ N }_{ 2 }\) evolved \({ V }_{ t } = 19.3\) $$ \therefore k = \frac { 2.303 }{ 6 } \log { \frac { 58.3 }{ 58.3 - 19.3 } } = 0.384\times \log { \left( 1.495 \right) } = 0.067{ min }^{ -1 }$$
at time t = 12 min volume of \({ N }_{ 2 }\) evolved \({ V }_{ t } = 32.6\) $$ \therefore k = \frac { 2.303 }{ 12 } \log { \frac { 58.3 }{ 58.3 - 32.6 } } = 0.192\times \log { \left( 2.268 \right) } = 0.068{ min }^{ -1 }$$
at time t = 18 min volume of \({ N }_{ 2 }\) evolved \({ V }_{ t } = 41.3\) $$ \therefore k = \frac { 2.303 }{ 18 } \log { \frac { 58.3 }{ 58.3 - 41.3 } } = 0.128\times \log { \left( 3.429 \right) } = 0.069{ min }^{ -1 }$$
at time t = 24 min volume of \({ N }_{ 2 }\) evolved \({ V }_{ t } = 46.5\) $$ \therefore k = \frac { 2.303 }{ 24 } \log { \frac { 58.3 }{ 58.3 - 46.5 } } = 0.096\times \log { \left( 4.941 \right) } = 0.067{ min }^{ -1 }$$
at time t = 30 min volume of \({ N }_{ 2 }\) evolved \({ V }_{ t } = 50.4\) $$ \therefore k = \frac { 2.303 }{ 30 } \log { \frac { 58.3 }{ 58.3 - 50.4 } } = 0.077\times \log { \left( 7.380 \right) } = 0.067{ min }^{ -1 }$$ Since the value of k comes out to be nearly constant, given reaction is of the first order. The average value of the rate constant = 0.067 \({ min }^{ -1 }\).
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