The initial concentration of N 2 O 5 in the following first order reaction, N 2 O 5(g) longrightarrow 2 NO 2(g)+1 / 2 O 2(g) was 1.24 × 10-2 mol L -1 at 318 K. The concentration of N 2 O 5 after 60 minutes was 0.20 × 10-2 mol L -1. The rate constant of the reaction at 318 K is

Question

The initial concentration of N 2 O 5 in the following first order reaction, N 2 O 5(g) longrightarrow 2 NO 2(g)+1 / 2 O 2(g) was 1.24 × 10-2 mol L -1 at 318 K. The concentration of N 2 O 5 after 60 minutes was 0.20 × 10-2 mol L -1. The rate constant of the reaction at 318 K is (A) k = 0.004 min-1  (B) k = 0.0304 min-1  (C)  k = 2.3 min-1  (D)  k = 1.5 min-1

Tardigrade Admin · Accepted Answer

For a first order reaction log ([R]1/[R]2)=(k(t2-t1)/2.303) k=(2.303/(t2-t1)) log ([R]1/[R]2) =(2.303/(60 min -0 min )) log (1.24 × 10-2 mol L -1/0.20 × 10-2 mol L -1) =(2.303/60) log 6.2 min -1 k=0.0304 min -1

$k = 0.004 mi n^{- 1}$

$k = 0.0304 mi n^{- 1}$

$k = 2.3 mi n^{- 1}$

$k = 1.5 mi n^{- 1}$

Q. The initial concentration of N2​O5​ in the following first order reaction, N2​O5(g)​⟶2NO2(g)​+1/2O2(g)​ was 1.24×10−2molL−1 at 318K. The concentration of N2​O5​ after 60 minutes was 0.20×10−2molL−1. The rate constant of the reaction at 318K is

k=0.004min−1

k=0.0304min−1

k=2.3min−1

k=1.5min−1

For a first order reaction log[R]2​[R]1​​=2.303k(t2​−t1​)​ k=(t2​−t1​)2.303​log[R]2​[R]1​​ =(60min−0min)2.303​log0.20×10−2molL−11.24×10−2molL−1​ =602.303​log6.2min−1 k=0.0304min−1

$k = 0.004 mi n^{- 1}$

$k = 0.0304 mi n^{- 1}$

$k = 2.3 mi n^{- 1}$

$k = 1.5 mi n^{- 1}$