320167 Consider the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) as
\(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\dfrac{1}{2} \mathrm{O}_{2}\)
The rate of reaction is given by
\(\frac{{ - {\text{d}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]}}{{{\text{dt}}}} = \frac{1}{2}\frac{{{\text{d}}\left[ {{\text{N}}{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = 2\frac{{{\text{d}}\left[ {{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = {{\text{k}}_{\text{1}}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
Therefore, \(\frac{{ - {\text{d}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_5}} \right]}}{{{\text{dt}}}} = {{\text{k}}_{\text{1}}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
\(\frac{{{\text{ + d}}\left[ {{\text{N}}{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = 2{{\text{k}}_{\text{1}}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right] = {\text{k}}_1^\prime \left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
\(\frac{{{\text{ + d}}\left[ {{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = \frac{1}{2}{{\text{k}}_1}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right] = {\text{k}}_1^{\prime \prime }\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
Choose the correct option.

320168 In a reaction, \(A+B \rightarrow\) Product, rate is doubled when the concentration of \(B\) is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and \(\mathrm{B})\) are doubled, rate law for the reaction can be written as

320169 The rate law for the reaction below is given by the expression, \({\text{r = k[A][B]}}\)
\({\text{A + B}} \to {\text{ Product}}\)
If the concentration of B is increased from 0.1 to 0.3 mole, keeping the value of A at 0.1 mole, the rate constant will be :

320170 For the reaction \({\rm{A + B}} \to {\rm{C}}\); starting with different initial concentrations of A and B, initial rate of reaction were determined graphically in four experiments.
Rate law of the reaction from above data is

320167 Consider the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) as
\(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\dfrac{1}{2} \mathrm{O}_{2}\)
The rate of reaction is given by
\(\frac{{ - {\text{d}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]}}{{{\text{dt}}}} = \frac{1}{2}\frac{{{\text{d}}\left[ {{\text{N}}{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = 2\frac{{{\text{d}}\left[ {{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = {{\text{k}}_{\text{1}}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
Therefore, \(\frac{{ - {\text{d}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_5}} \right]}}{{{\text{dt}}}} = {{\text{k}}_{\text{1}}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
\(\frac{{{\text{ + d}}\left[ {{\text{N}}{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = 2{{\text{k}}_{\text{1}}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right] = {\text{k}}_1^\prime \left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
\(\frac{{{\text{ + d}}\left[ {{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = \frac{1}{2}{{\text{k}}_1}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right] = {\text{k}}_1^{\prime \prime }\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
Choose the correct option.

320168 In a reaction, \(A+B \rightarrow\) Product, rate is doubled when the concentration of \(B\) is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and \(\mathrm{B})\) are doubled, rate law for the reaction can be written as

320169 The rate law for the reaction below is given by the expression, \({\text{r = k[A][B]}}\)
\({\text{A + B}} \to {\text{ Product}}\)
If the concentration of B is increased from 0.1 to 0.3 mole, keeping the value of A at 0.1 mole, the rate constant will be :

320170 For the reaction \({\rm{A + B}} \to {\rm{C}}\); starting with different initial concentrations of A and B, initial rate of reaction were determined graphically in four experiments.
Rate law of the reaction from above data is

320167 Consider the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) as
\(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\dfrac{1}{2} \mathrm{O}_{2}\)
The rate of reaction is given by
\(\frac{{ - {\text{d}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]}}{{{\text{dt}}}} = \frac{1}{2}\frac{{{\text{d}}\left[ {{\text{N}}{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = 2\frac{{{\text{d}}\left[ {{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = {{\text{k}}_{\text{1}}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
Therefore, \(\frac{{ - {\text{d}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_5}} \right]}}{{{\text{dt}}}} = {{\text{k}}_{\text{1}}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
\(\frac{{{\text{ + d}}\left[ {{\text{N}}{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = 2{{\text{k}}_{\text{1}}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right] = {\text{k}}_1^\prime \left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
\(\frac{{{\text{ + d}}\left[ {{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = \frac{1}{2}{{\text{k}}_1}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right] = {\text{k}}_1^{\prime \prime }\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
Choose the correct option.

320168 In a reaction, \(A+B \rightarrow\) Product, rate is doubled when the concentration of \(B\) is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and \(\mathrm{B})\) are doubled, rate law for the reaction can be written as

320169 The rate law for the reaction below is given by the expression, \({\text{r = k[A][B]}}\)
\({\text{A + B}} \to {\text{ Product}}\)
If the concentration of B is increased from 0.1 to 0.3 mole, keeping the value of A at 0.1 mole, the rate constant will be :

320170 For the reaction \({\rm{A + B}} \to {\rm{C}}\); starting with different initial concentrations of A and B, initial rate of reaction were determined graphically in four experiments.
Rate law of the reaction from above data is

320167 Consider the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) as
\(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\dfrac{1}{2} \mathrm{O}_{2}\)
The rate of reaction is given by
\(\frac{{ - {\text{d}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]}}{{{\text{dt}}}} = \frac{1}{2}\frac{{{\text{d}}\left[ {{\text{N}}{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = 2\frac{{{\text{d}}\left[ {{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = {{\text{k}}_{\text{1}}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
Therefore, \(\frac{{ - {\text{d}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_5}} \right]}}{{{\text{dt}}}} = {{\text{k}}_{\text{1}}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
\(\frac{{{\text{ + d}}\left[ {{\text{N}}{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = 2{{\text{k}}_{\text{1}}}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right] = {\text{k}}_1^\prime \left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
\(\frac{{{\text{ + d}}\left[ {{{\text{O}}_{\text{2}}}} \right]}}{{{\text{dt}}}} = \frac{1}{2}{{\text{k}}_1}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right] = {\text{k}}_1^{\prime \prime }\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right]\)
Choose the correct option.

320168 In a reaction, \(A+B \rightarrow\) Product, rate is doubled when the concentration of \(B\) is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and \(\mathrm{B})\) are doubled, rate law for the reaction can be written as

320169 The rate law for the reaction below is given by the expression, \({\text{r = k[A][B]}}\)
\({\text{A + B}} \to {\text{ Product}}\)
If the concentration of B is increased from 0.1 to 0.3 mole, keeping the value of A at 0.1 mole, the rate constant will be :

320170 For the reaction \({\rm{A + B}} \to {\rm{C}}\); starting with different initial concentrations of A and B, initial rate of reaction were determined graphically in four experiments.
Rate law of the reaction from above data is