A) Its order is 1 and molecularity is 1
B) Its order is 1 and molecularity is 2
C) Its order is 2 and molecularity is 2
D) Its order is 2 and molecularity is 1
Correct Answer: A
Solution :
\[2{{N}_{2}}{{O}_{5}}(g)\to 4N{{O}_{2}}(g)+{{O}_{2}}(g);\] \[\text{Rate}=k[{{N}_{2}}{{O}_{5}}]\] The reaction occurs in two steps as Step 1 \[{{N}_{2}}{{O}_{5}}\xrightarrow{\text{slow}}N{{O}_{2}}+N{{O}_{3}}\] Step 2 \[{{N}_{2}}{{O}_{5}}+N{{O}_{3}}\xrightarrow{fast}3N{{O}_{2}}+{{O}_{2}}\] The slow step is unimolecular while fast step is bimolecular. Hence, the above reaction is unimolecular and its order is 1 (first).
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