Question

58 The rate for the following to be 0.22M NO2 I min. w rate at 320K? ollowing reaction at book was found 102 1 min. what would be the approximate N204 Lg) - ZNO2Lg) the following rate St. A reaction occurs in three steps with constants: KI=0. 3M K2=0.05M, Step 1 (a) which step is the slow step? AK D 3M B K2=00SM ( +3=4.5MD Ck3=4.50 Step 2 Step 3 (b) which step has the lowest activation energy?

please help:) thank you

Answer 1

58.

Rate of reaction is almost doubled by every 10⁰ c or 10 K rise.

In the given problem change in temperature = (320 - 300) = 20K or 20⁰c

So, increasing 20⁰ c the rate of reaction is 4 times than initial rate.

So, approximate rate at 320 K = 4×0.22 = 0.88 M NO₂/min.

62.

a)

Rate of reaction is proportional to rate constant.

Step 2 has lowest value of rate constant ( K).

So, step 2 is the slow step.

b)

From arrhenious equation

K = A×e^-Ea/RT

Hence, higher the rate constant , lower the value of activation energy

Step 3 has the highest value of rate constant.

Hence, step 3 has the lowest activation energy.