The specific rate constant of the decomposition of N2O5 is 0.008 min-1. The volume of O2 collected after 20 minutes is 16 mL. Find the volume that would be collected at the end of reaction. NO2 formed is dissolved in CCl4.

Question

The specific rate constant of the decomposition of N2O5 is 0.008 min-1. The volume of O2 collected after 20 minutes is 16 mL. Find the volume that would be collected at the end of reaction. NO2 formed is dissolved in CCl4. (A) 17.49 mL (B) 27.49 mL (C) 37.49 mL (D) 47.49 mL

Tardigrade Admin · Accepted Answer

Given k = 0.008 min-1 From unit of k, the reaction is of the order of first. From ∴ k= (2.303/t) log (V∞/V∞ -Vt) ⇒ 0.008 = (2.303/20) log (V∞/V∞ - 16) ⇒ 0.0695 = log (V∞/V∞ - 16) ⇒ V∞ = 17.49 mL

Q. The specific rate constant of the decomposition of $N_{2} O_{5}$ is $0.008$ min $^{- 1}$ . The volume of $O_{2}$ collected after $20$ minutes is $16 m L$ . Find the volume that would be collected at the end of reaction. $N O_{2}$ formed is dissolved in $CC l_{4}$ .

$17.49 m L$

$27.49 m L$

$37.49 m L$

$47.49 m L$

Q. The specific rate constant of the decomposition of N2​O5​ is 0.008 min−1. The volume of O2​ collected after 20 minutes is 16mL. Find the volume that would be collected at the end of reaction. NO2​ formed is dissolved in CCl4​.

17.49mL

27.49mL

37.49mL

47.49mL

Given k=0.008 min−1 From unit of k, the reaction is of the order of first. From ∴k=t2.303​ log V∞​−Vt​V∞​​ ⇒0.008=202.303​ log V∞​−16V∞​​ ⇒0.0695= log V∞​−16V∞​​ ⇒V∞​=17.49mL

Q. The specific rate constant of the decomposition of $N_{2} O_{5}$ is $0.008$ min $^{- 1}$ . The volume of $O_{2}$ collected after $20$ minutes is $16 m L$ . Find the volume that would be collected at the end of reaction. $N O_{2}$ formed is dissolved in $CC l_{4}$ .

$17.49 m L$

$27.49 m L$

$37.49 m L$

$47.49 m L$