Gaseous N2O5 decomposes according to the following equation: N2O5(g)→2NO2(g)+(1/2)O2(g) The experimental rate law is-Δ[N2O5]/Δ t=k[N2O5] At a certain temperature the rate constant is k = 5.0 × 10-4 sec-1. In how seconds will the concentration of N2O5 decrease to one-tenth of its initial value?

Question

Gaseous N2O5 decomposes according to the following equation: N2O5(g)→2NO2(g)+(1/2)O2(g) The experimental rate law is-Δ[N2O5]/Δ t=k[N2O5] At a certain temperature the rate constant is k = 5.0 × 10-4 sec-1. In how seconds will the concentration of N2O5 decrease to one-tenth of its initial value? (A) 2.0 × 103 s (B) 4.6 × 103s (C) 2.1 × 102 s (D) 1.4 × 103 s

Tardigrade Admin · Accepted Answer

t=(2.303/5×10-4) log(A0/A0/10)=4.6×103sec

$2.0 \times 1 0^{3} s$

$4.6 \times 1 0^{3} s$

$2.1 \times 1 0^{2} s$

$1.4 \times 1 0^{3} s$

$t = \frac{2.303}{5 \times 1 0 ^{- 4}} l o g \frac{A _{0}}{A _{0} /10} = 4.6 \times 1 0^{3} sec$

2.0×103s

4.6×103s

2.1×102s

1.4×103s

t=5×10−42.303​logA0​/10A0​​=4.6×103sec

$2.0 \times 1 0^{3} s$

$4.6 \times 1 0^{3} s$

$2.1 \times 1 0^{2} s$

$1.4 \times 1 0^{3} s$

$t = \frac{2.303}{5 \times 1 0 ^{- 4}} l o g \frac{A _{0}}{A _{0} /10} = 4.6 \times 1 0^{3} sec$