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Published on: 28/05/2021
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Questions + Answers key
Take MCQ Chemistry Test1.
50% of the original amount of a reactant was added to the reaction mixture after 40 min. What % of the total amount will be present after 60 min, given that half life period of the reaction is 20 min.
2.
A heterogenous reaction is carried out at 500 K. If the same reaction is carried out in the presence of catalyst at the same rate, the temperature requires is 400 K, calculate the activation energy of the reaction if the catalyst lowers the activation barrier by 20 KJ/mol.
3.
The gas phase decomposition of CH3OCH3 follows first order kinetics
CH3OCH3 ⟶ CH4(g) + H2(g) + CO(g)
The reaction is carried out in a constant volume container at 5000 C and has t1/2 = 14.5 min. Initially only dimethyl ether is present at a pressure of 0.40 atm. What is the total pressure of the system after 12 min? Assume ideal behavior.
4.
For the reaction, the energy of activation is 75KJ / mol. When a catalyst is added the reaction its energy of activation is lowered to 20KJ / mol. What is the effect of catalyst on the rate of reaction at 200C.
1.
40 min = 2 half lives
Amount present after 40 min = a/4
Amount added = a/2
Total amount after 40 min = a/4 + a/2 = 3a/4
After 60 min from start , amount left = 3a/8
Total amount taken = a + a/2 = 3a/2
Hence % = 3a/8 ÷ 3a/2 x 100 = 25%.
2.
K = Ae-Ea/RT
Ea, absence of catalyst = x KJ/mol
Ea, presence of catalyst = (x-20)KJ/mol
At 500K K = Ae-x/R.500
At 400
K = Ae-(x-20)/R.400
Divide 1 by 2, we get
X = 100KJ/mol
Ea = 100KJ/mol.
3.
K = 0.693/t1/2
(a-x) ∝ (0.40 –P)atm.
K = 2.303/12 log 0.40/0.40 –P
Pcalculated = 0.1745 atm
Total pressure = 0.4 -P+P+P+P = 0.749 atm
4.
Log K = log A - Ea/2.303RT
Log K’ = log A - Ea/2.303RT
Log K = log A - Ea/2.303RT
Log K’/K = Ea-Ea/2.303 RT = 9.8037
Or K’/K = 6.36 x 109
UR catalysted /UR uncatalysed = K’/K = 6.4 x 10 9
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