The rate for the reaction 2N2O5 (. g .) arrow 4(NO)2 (. g .)+O2 (. g .) is 2.4× 10- 5mollit- 1sec- 1. If the rate is 3.0× 10- 5 sec- 1 then the concentration of N2O5 in mollit- 1 is: (Fill your anwer by multiply with 10)

Question

The rate for the reaction 2N2O5 (. g .) arrow 4(NO)2 (. g .)+O2 (. g .) is 2.4× 10- 5mollit- 1sec- 1. If the rate is 3.0× 10- 5 sec- 1 then the concentration of N2O5 in mollit- 1 is: (Fill your anwer by multiply with 10) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Rate =k[N2 O5] Hence 2.4× 10- 5=3.0× 10- 5IN2O5l or [N2 O5]=0.8molL- 1

Q. The rate for the reaction $2 N_{2} O_{5 (g)} \to 4 (NO)_{2 (g)} + O_{2 (g)}$ is $2.4 \times 1 0^{- 5} m o ll i t^{- 1} se c^{- 1} .$ If the rate is $3.0 \times 1 0^{- 5}$ $se c^{- 1}$ then the concentration of $N_{2} O_{5}$ in $m o ll i t^{- 1}$ is:
(Fill your anwer by multiply with 10)

Rate $= k [N_{2} O_{5}]$
Hence $2.4 \times 1 0^{- 5} = 3.0 \times 1 0^{- 5} I N_{2} O_{5} l$
or $[N_{2} O_{5}] = 0.8 m o l L^{- 1}$

Q. The rate for the reaction 2N2​O5(g)​→4(NO)2(g)​+O2(g)​ is 2.4×10−5mollit−1sec−1. If the rate is 3.0×10−5 sec−1 then the concentration of N2​O5​ in mollit−1 is: (Fill your anwer by multiply with 10)

Rate =k[N2​O5​] Hence 2.4×10−5=3.0×10−5IN2​O5​l or [N2​O5​]=0.8molL−1

Q. The rate for the reaction $2 N_{2} O_{5 (g)} \to 4 (NO)_{2 (g)} + O_{2 (g)}$ is $2.4 \times 1 0^{- 5} m o ll i t^{- 1} se c^{- 1} .$ If the rate is $3.0 \times 1 0^{- 5}$ $se c^{- 1}$ then the concentration of $N_{2} O_{5}$ in $m o ll i t^{- 1}$ is:
(Fill your anwer by multiply with 10)

Rate $= k [N_{2} O_{5}]$
Hence $2.4 \times 1 0^{- 5} = 3.0 \times 1 0^{- 5} I N_{2} O_{5} l$
or $[N_{2} O_{5}] = 0.8 m o l L^{- 1}$