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Hint: We know that rate of reaction can be calculated from the concentration of reactants if the rate law of the reaction is known. We can also calculate the rate of reaction from the rate of decomposition of reactants and the rate of formation of products.
Complete step by step answer: We know that we calculate the reaction rate from the differential terms of the reactants. The rate can be expressed in terms of the rate of decomposition of reactants and terms of rate of formation of products.
Rate of reaction r = $-\dfrac { 1 }{ 2 } \dfrac { d\left[ { N }_{ 2 }{ O }_{ 5 } \right] }{ dt } = \dfrac { 1 }{ 4 } \dfrac { d\left[ { NO }_{ 2 } \right] }{ dt } = \dfrac { d\left[ { O }_{ 2 } \right] }{ dt } $
Concentration of $N_{ 2 }O_{ 5 }$,$NO_{ 2 }$,$O_{ 2 }$ at t=0 in$molL^{ -1 }$ are 7.2, 0, 0 respectively
Concentration of $N_{ 2 }O_{ 5 }$,$NO_{ 2 }$,$O_{ 2 }$ at t=t in$molL^{ -1 }$ are 7.2-2x, 4x, x respectively
Given the concentration of $NO_{ 2 }$ is 2.4 $molL^{ -1 }$ at time t=20min.
Equate t with 20 then $4x = 2.4$
\[x = \dfrac { 2.4 }{ 4 } = 0.6molL^{ -1 }\]
Now we can calculate the concentration of $N_{ 2 }O_{ 5 }$.
\[[N_{ 2 }O_{ 5 }] in molL^{ -1 } = 7.2 - 2x = 7.2 - 2(0.6) = 7.2 - 1.2 = 6\]
Rate of decomposition of $N_{ 2 }O_{ 5 }$ =
\[\dfrac { \left[ { N }_{ 2 }{ O }_{ 5 } \right] _{ t=0 } - { \left[ { N }_{ 2 }{ O }_{ 5 } \right] }_{ t=20 } }{ 20 - 0 } = \dfrac { \left( 7.2 - 6 \right) { molL }^{ -1 } }{ 20 min } = \dfrac { 1.2 }{ 20 } mol{ L }^{ -1 } min^{ -1 }\]
Rate of decomposition of $N_{ 2 }O_{ 5 }$ = 0.06 $molL^{ -1 } min^{ -1 }$
Therefore, option A is correct.
Note: Take the decomposition of the reactant correctly. Check the sign correctly. The rate should be written by taking stoichiometric coefficients in the reaction correctly. Time should be taken in minutes as we are asked to answer given units.
Complete step by step answer: We know that we calculate the reaction rate from the differential terms of the reactants. The rate can be expressed in terms of the rate of decomposition of reactants and terms of rate of formation of products.
Rate of reaction r = $-\dfrac { 1 }{ 2 } \dfrac { d\left[ { N }_{ 2 }{ O }_{ 5 } \right] }{ dt } = \dfrac { 1 }{ 4 } \dfrac { d\left[ { NO }_{ 2 } \right] }{ dt } = \dfrac { d\left[ { O }_{ 2 } \right] }{ dt } $
Concentration of $N_{ 2 }O_{ 5 }$,$NO_{ 2 }$,$O_{ 2 }$ at t=0 in$molL^{ -1 }$ are 7.2, 0, 0 respectively
Concentration of $N_{ 2 }O_{ 5 }$,$NO_{ 2 }$,$O_{ 2 }$ at t=t in$molL^{ -1 }$ are 7.2-2x, 4x, x respectively
Given the concentration of $NO_{ 2 }$ is 2.4 $molL^{ -1 }$ at time t=20min.
Equate t with 20 then $4x = 2.4$
\[x = \dfrac { 2.4 }{ 4 } = 0.6molL^{ -1 }\]
Now we can calculate the concentration of $N_{ 2 }O_{ 5 }$.
\[[N_{ 2 }O_{ 5 }] in molL^{ -1 } = 7.2 - 2x = 7.2 - 2(0.6) = 7.2 - 1.2 = 6\]
Rate of decomposition of $N_{ 2 }O_{ 5 }$ =
\[\dfrac { \left[ { N }_{ 2 }{ O }_{ 5 } \right] _{ t=0 } - { \left[ { N }_{ 2 }{ O }_{ 5 } \right] }_{ t=20 } }{ 20 - 0 } = \dfrac { \left( 7.2 - 6 \right) { molL }^{ -1 } }{ 20 min } = \dfrac { 1.2 }{ 20 } mol{ L }^{ -1 } min^{ -1 }\]
Rate of decomposition of $N_{ 2 }O_{ 5 }$ = 0.06 $molL^{ -1 } min^{ -1 }$
Therefore, option A is correct.
Note: Take the decomposition of the reactant correctly. Check the sign correctly. The rate should be written by taking stoichiometric coefficients in the reaction correctly. Time should be taken in minutes as we are asked to answer given units.
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