The decomposition of dinitrogen pentoxide (N2O5) follows first order rate law. What will be the rate constant from the given data? At t = 800 s, [N2O5] = 1.45 mol L-1 At t = 1600 s, [N2O5] = 0.88 mol L-1

Question

The decomposition of dinitrogen pentoxide (N2O5) follows first order rate law. What will be the rate constant from the given data? At t = 800 s, [N2O5] = 1.45 mol L-1 At t = 1600 s, [N2O5] = 0.88 mol L-1 (A) 3.12 × 10-4 s -1 (B) 6.24 × 10-4 s -1 (C) 2.84 × 10-4 s -1 (D) 8.14 × 10-4 s -1

Tardigrade Admin · Accepted Answer

k=(2.303/(t2-t1)) log ([A1]/[A2]) k=(2.303/(1600-800)) log (1.45/0.88) =(2.303/800)×0.2169 =6.24×10-4 s-1

Q. The decomposition of dinitrogen pentoxide $(N_{2} O_{5})$ follows first order rate law. What will be the rate constant from the given data?
At $t = 800 s$ , $[N_{2} O_{5}]$ $= 1.45 m o l$ $L^{- 1}$
At $t = 1600 s$ , $[N_{2} O_{5}]$ $= 0.88 m o l$ $L^{- 1}$

$3.12 \times 1 0^{- 4}$ $s^{- 1}$

$6.24 \times 1 0^{- 4}$ $s^{- 1}$

$2.84 \times 1 0^{- 4}$ $s^{- 1}$

$8.14 \times 1 0^{- 4}$ $s^{- 1}$

$k = \frac{2.303}{( t _{2} - t _{1} )} l o g$ $\frac{[ A _{1} ]}{[ A _{2} ]}$
$k = \frac{2.303}{( 1600 - 800 )} l o g$ $\frac{1.45}{0.88}$ $= \frac{2.303}{800} \times 0.2169$
$= 6.24 \times 1 0^{- 4}$ $s^{- 1}$

Q. The decomposition of dinitrogen pentoxide (N2​O5​) follows first order rate law. What will be the rate constant from the given data? At t=800s, [N2​O5​] =1.45mol L−1 At t=1600s, [N2​O5​] =0.88mol L−1

3.12×10−4 s−1

6.24×10−4 s−1

2.84×10−4 s−1

8.14×10−4 s−1