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Question: For the reaction \(2N_{2}O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}\), if concentration of \(NO_{2}...

For the reaction 2N2O5(g)→4NO2(g)+O2(g)2N_{2}O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}, if

concentration of NO2NO_{2} in 100 seconds is increased by 5.2×10−3m5.2 \times 10^{- 3}m. Then rate of reaction will be.

A

1.3×10−5ms−11.3 \times 10^{- 5}ms^{- 1}

B

5×10−4ms−15 \times 10^{- 4}ms^{- 1}

C

7.6×10−4ms−17.6 \times 10^{- 4}ms^{- 1}

D

2×10−3ms−12 \times 10^{- 3}ms^{- 1}

(5) 2.5×10−5ms−12.5 \times 10^{- 5}ms^{- 1}

Answer

1.3×10−5ms−11.3 \times 10^{- 5}ms^{- 1}

Explanation

Solution

2N2O5(g)→4NO2(g)+O2(g)2N_{2}O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}

Rate of reaction with respect to NO2NO_{2}

=14d[NO2]dt=14×5.2×10−3100=1.3×10−5ms−1= \frac{1}{4}\frac{d\lbrack NO_{2}\rbrack}{dt} = \frac{1}{4} \times \frac{5.2 \times 10^{- 3}}{100} = 1.3 \times 10^{- 5}ms^{- 1}