Based on their activation energies and energy changes and assuming that all coll

Question

Based on their activation energies and energy changes and assuming that all collision factors are the same, which of the following reactions would be slowest in the forward direction? E_a = 322; delta E = 218; E_a = 128; delta E = 213; E_a = 235; delta E = 189; E_a = 76; delta E = -133; E_a = 161; delta E = -162; For the elementary processes whose equations are given below, determine whether each is unimolecular or bimolecular and determine the rate law for each process. Pick the choice which is correct. NOCI_2(g) rightarrow NO(g) + CI_2(g) is bimolecular, with rate = k [NOCI_2]^2. NO(g) + CI_2(g) rightarrow NOCI_2(g) is unimolecular, with rate = k[NO] [CI_2]/[NOCI_2]. 2 CI(g) rightarrow CI_2(g) is unimolecular, with rate = k [CI]. CI_2(g) rightarrow 2 CI(g) is unimolecular, with rate = K [CI_2]. CI(g) + HBr(g) rightarrow HCI(g) + Br(g) is bimolecular, with rate = k [HCI] [Br]. The reaction, Co_2(g) + 2 OH^-(aq) rightarrow CO_3^2- (aq) + H_2O(I), Proceeds through the following mechanism: CO_2(aq) + OH^-(aq) rightarrow HCO_3^-(aq) HCO_3^-(aq) + OH^-(aq) rightarrow CO_3^2- (aq) + H_2O(I) The first step of this mechanism is rate-determining (slow). What would the rate law fo reaction be if the second step were rate-determining? Rate = k [OH^-]^2 [CO_2]^1/2 Rate = k [OH^-] [CO_2]^1/2 Rate = k [OH^-]^2 [CO_2] Rate = k [OH^-]^2 Rate = k [OH^-]^1/2 [CO_2}

Explanation / Answer

13)
K = A*e^(-Ea/RT)
so,
greater the value of Ea, smaller will be rate constant
Hence the reaction with highest Ea value will be slowest
Answer: a

14)
a is unimolecular
b is bi molecular
c is bi molecular
d is unimolecular and rate expression is also correct
e is bi molecular but rate expression is wrong
Answer: d

I am allowed to answer only 1 question at a time

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