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Question: The decomposition of di-nitrogen pentoxide \((N_{2}O_{5})\) follows first order rate law. What will ...

The decomposition of di-nitrogen pentoxide (N2O5)(N_{2}O_{5}) follows first order rate law. What will be the rate constant from the given data?

At t = 800 s [N2O5]=1.45molL−1\lbrack N_{2}O_{5}\rbrack = 1.45molL^{- 1}

At t = 1600 s, [N2O5]=0.88molL−1\lbrack N_{2}O_{5}\rbrack = 0.88molL^{- 1}

A

3.12×10−4s−13.12 \times 10^{- 4}s^{- 1}

B

6.24×10−4s−16.24 \times 10^{- 4}s^{- 1}

C

2.84×10−4s−12.84 \times 10^{- 4}s^{- 1}

D

8.14×10−4s−18.14 \times 10^{- 4}s^{- 1}

Answer

6.24×10−4s−16.24 \times 10^{- 4}s^{- 1}

Explanation

Solution

k=2.303(t2−t1)log⁡[A1][A2]k = \frac{2.303}{(t_{2} - t_{1})}\log\frac{\lbrack A_{1}\rbrack}{\lbrack A_{2}\rbrack}

k=2.303(1600−800)log⁡1.450.88=2.303800×0.2169k = \frac{2.303}{(1600 - 800)}\log\frac{1.45}{0.88} = \frac{2.303}{800} \times 0.2169

=6.24×10−4s−1= 6.24 \times 10^{- 4}s^{- 1}