NEET Chemistry Chemical Kinetics Questions Solved


1 mole of gas changes linearly from initial state (2 atm, 10 lt) to final state (8 atm, 4 lt). Find the value of rate constant, at the maximum temperature, that the gas can attain. Maximum rate constant is equal to 20 sec-1 and value of activation energy is 40 kJ mole-1, assuming that activation energy does not change in this temperature range.

(A) 0.56 × 10-3 sec-1                                          

(B) 3.16 × 10-3 sec-1

(C) 1.56 × 10-3sec-1                                          

(D) 5.12 × 10-3 sec-1

(C) Let the equation of straight line is P=mV+c

  Now putting the values, we get P+V=12

From the ideal gas equation,

T=PVnR=PVR; T=12-VVR

For T to be maximum dTdV=0

V=6 lt, P=6 atm.

Value of Tmax=360.082=439 K

Now Putting the values of A=20 sec-1

Ea=40×103J mole-1

we get, k=A e-Ea/RT

k=20 e-40×1038.314×439 ; k=1.56×10-3sec-1

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