The reaction, N2O5 in CCl4 text(solution)arrow 2NO2 text(solution)+(1/2)O2(g) is of first order in N2O5 with rate constant 6.2× 10-4s-1. What is the value of rate of reaction when [N2O5]=1.25 mol L-1 ?

Question

The reaction, N2O5 in CCl4 text(solution)arrow 2NO2 text(solution)+(1/2)O2(g) is of first order in N2O5 with rate constant 6.2× 10-4s-1. What is the value of rate of reaction when [N2O5]=1.25 mol L-1 ? (A) 5.15 × 10-5 mol L -1 s -1 (B) 6.35 × 10-3 mol L -1 s -1 (C) 7.75 × 10-4 mol L -1 s -1 (D) 3.85 × 10-4 mol L -1 s -1

Tardigrade Admin · Accepted Answer

For first order reaction, Rate =k[ N 2 O 5] =6.2 × 10-4 s -1 × 1.25 mol L -1 =7.75 × 10-4 mol L -1 s -1

Q. The reaction, $N_{2} O_{5}$ in $CC l_{4} (solution) \to 2 N O_{2} (solution) + \frac{1}{2} O_{2} (g)$
is of first order in $N_{2} O_{5}$ with rate constant $6.2 \times 1 0^{- 4} s^{- 1} .$
What is the value of rate of reaction when $[N_{2} O_{5}] = 1.25 m o l L^{- 1}$ ?

$5.15 \times 1 0^{- 5} m o l L^{- 1} s^{- 1}$

$6.35 \times 1 0^{- 3} m o l L^{- 1} s^{- 1}$

$7.75 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}$

$3.85 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}$

For first order reaction,
Rate $= k [N_{2} O_{5}]$
$= 6.2 \times 1 0^{- 4} s^{- 1} \times 1.25 m o l L^{- 1}$
$= 7.75 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}$

Q. The reaction, N2​O5​ in CCl4​(solution)→2NO2​(solution)+21​O2​(g) is of first order in N2​O5​ with rate constant 6.2×10−4s−1. What is the value of rate of reaction when [N2​O5​]=1.25molL−1 ?

5.15×10−5molL−1s−1

6.35×10−3molL−1s−1

7.75×10−4molL−1s−1

3.85×10−4molL−1s−1