For the decomposition of N$_2$O$_5$ at 298 K, the rate law is $-\dfrac{d[\mathrm{N_2O_5}]}{dt}=k[\mathrm{N_2O_5}]$. Starting with 2.5 moles of N$_2$O$_5$(g) in a five-litre container, how many moles per litre of N$_2$O$_5$ would remain after 75 seconds if the rate constant for the reaction is $16.8\times10^{-3}$ s$^{-1}$?

CBSE Class 12 · 2000 · All India Compartment · Chemical KineticsShort Answer

For the decomposition of N

_{2}

_{5}

at 298 K, the rate law is

- \frac{d [ N _{2} O _{5} ]}{d t} = k [N_{2} O_{5}]

. Starting with 2.5 moles of N

_{2}

_{5}

(g) in a five-litre container, how many moles per litre of N

_{2}

_{5}

would remain after 75 seconds if the rate constant for the reaction is

16.8 \times 1 0^{- 3}

^{- 1}

Answer: ≈ 0.142 mol L⁻¹

Initial concentration:

[R]_{0} = \frac{2.5}{5} = 0.5 mol L^{- 1}

First order integrated equation:

lo g \frac{[ R ] _{0}}{[ R ]} = \frac{k t}{2.303} = \frac{16.8 \times 1 0 ^{- 3} \times 75}{2.303} = 0.547

\frac{[ R ] _{0}}{[ R ]} = 3.52

[R] = \frac{0.5}{3.52}

[R] = 0.142 mol L^{- 1}

CBSE Class 12 Board · 2000 · All India Compartment · Subjective