Chapter 17 – Chemical Kinetics
Sahay Sir > Question Answers > AIIMS Last 24 Years Solved 1996 - 2019 Physics and Chemistry Video Solutions > Chemistry > Chapter 17 - Chemical Kinetics
If the rate of decomposition of N2O5 during a certain time internal is 2.4×10^−4 mol L^−1 min^−1. N2O5 →2NO2 + 2/1O2 What is the rate of formation of NO2 and O2 mol L^−1 min^−1 ?
02
Jun
If the rate of decomposition of N2O5 during a certain time internal is 2.4×10^−4 mol L^−1 min^−1. N2O5 →2NO2 + 2/1O2 What is the rate of formation of NO2 and O2 mol L^−1 min^−1 ? If the rate of decomposition of N2O5 during a certain time internal is 2.4×10^−4 mol L^−1 min^−1. N2O5 →2NO2 + [...]
In the reaction A gives product , -da/dt = k1A. If we start with 10 M of A, then after one natural life time, concentration of A is decreased to
02
Jun
In the reaction A gives product , -da/dt = k1A. If we start with 10 M of A, then after one natural life time, concentration of A is decreased to -da/dt = k1A. If we start with 10 M of A concentration of A is decreased to In the reaction A gives product then after [...]
For the reaction, 2NO + Br2 → 2NOBr following mechanism has been given: (1) NO+Br2 ⇌Fast NOBr2 (2) NOBr2 +NO gives Slow 2NOBr Hence, rate law is
02
Jun
For the reaction, 2NO + Br2 → 2NOBr following mechanism has been given: (1) NO+Br2 ⇌Fast NOBr2 (2) NOBr2 +NO gives Slow 2NOBr Hence, rate law is 2NO + Br2 → 2NOBr following mechanism has been given: (1) NO+Br2 ⇌Fast NOBr2 (2) NOBr2 +NO gives Slow 2NOBr Hence For the reaction rate law is June [...]
For nth order reaction (dx/dt)=Rate = k[A]0n Graph between log (rate) against log[A0] is of the type Lines P, Q, R, S are of the order:
02
Jun
For nth order reaction (dx/dt)=Rate = k[A]0n Graph between log (rate) against log[A0] is of the type Lines P, Q, R, S are of the order: -q For nth order reaction (dx/dt)=Rate = k[A]0n Graph between log (rate) against log[A0] is of the type Lines P R S are of the order: the activation energy [...]
Rate constant k varies with temperature as given by equation log k (min^−1) = 5 − 2000/T Consider the following about this equation
02
Jun
Rate constant k varies with temperature as given by equation log k (min^−1) = 5 − 2000/T Consider the following about this equation Rate constant k varies with temperature as given by equation log k (min^−1) = 5 − 2000/T Consider the following about this equation June 2, 2021 Category: AIIMS Last 24 Years Solved [...]
On introducing a catalyst at 300 k, the activation energy increases by 5.75 kJ/mol. The ratio of rate constant k2/k1 will be
02
Jun
On introducing a catalyst at 300 k, the activation energy increases by 5.75 kJ/mol. The ratio of rate constant k2/k1 will be On introducing a catalyst at 300 k the activation energy increases by 5.75 kJ/mol. The ratio of rate constant k2/k1 will be June 2, 2021 Category: AIIMS Last 24 Years Solved 1996 - [...]