For the reaction, 2N2O5 → 4NO2 + O2, the rate of reaction can be expressed in terms of time and concentration by the expression:

Question

For the reaction, 2N2O5 → 4NO2 + O2, the rate of reaction can be expressed in terms of time and concentration by the expression: (A) Rate = -(d[N2O5]/dt)=-(1/4)(d[NO2]/dt)=(1/2)(d[O2]/dt) (B) Rate = -(1/2) (d[N2O5]/dt)=(1/4)(d[NO2]/dt)=(d[O2]/dt) (C) Rate = -(1/4) (d[N2O5]/dt)=(1/2)(d[NO2]/dt)=(d[O2]/dt) (D) Rate = -(1/2) (d[N2O5]/dt)=(1/2)(d[NO2]/dt)=(1/2) (d[O2]/dt)

Tardigrade Admin · Accepted Answer

Correct answer is (b) Rate = -(1/2) (d[N2O5]/dt)=(1/4)(d[NO2]/dt)=(d[O2]/dt)

Q. For the reaction, $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ , the rate of reaction can be expressed in terms of time and concentration by the expression:

Rate $= - \frac{d [ N _{2} O _{5} ]}{d t} = - \frac{1}{4} \frac{d [ N O _{2} ]}{d t} = \frac{1}{2} \frac{d [ O _{2} ]}{d t}$

Rate $= - \frac{1}{2} \frac{d [ N _{2} O _{5} ]}{d t} = \frac{1}{4} \frac{d [ N O _{2} ]}{d t} = \frac{d [ O _{2} ]}{d t}$

Rate $= - \frac{1}{4} \frac{d [ N _{2} O _{5} ]}{d t} = \frac{1}{2} \frac{d [ N O _{2} ]}{d t} = \frac{d [ O _{2} ]}{d t}$

Rate $= - \frac{1}{2} \frac{d [ N _{2} O _{5} ]}{d t} = \frac{1}{2} \frac{d [ N O _{2} ]}{d t} = \frac{1}{2} \frac{d [ O _{2} ]}{d t}$

Correct answer is (b) Rate $= - \frac{1}{2} \frac{d [ N _{2} O _{5} ]}{d t} = \frac{1}{4} \frac{d [ N O _{2} ]}{d t} = \frac{d [ O _{2} ]}{d t}$

Q. For the reaction, 2N2​O5​→4NO2​+O2​, the rate of reaction can be expressed in terms of time and concentration by the expression:

Rate =−dtd[N2​O5​]​=−41​dtd[NO2​]​=21​dtd[O2​]​

Rate =−21​dtd[N2​O5​]​=41​dtd[NO2​]​=dtd[O2​]​

Rate =−41​dtd[N2​O5​]​=21​dtd[NO2​]​=dtd[O2​]​

Rate =−21​dtd[N2​O5​]​=21​dtd[NO2​]​=21​dtd[O2​]​