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JEE chemistry problem

The reaction; N2O5 in 2NO2+12O2(g) is of first order for N2O5 with rate constant 6.2×10^−4 s^−1. What is the value of rate of reaction when [N2O5]=1.25 mol L−1?

Note by Ashley Shamidha
2 years, 6 months ago

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When SO2 is passed through a solution of potassium iodate, the oxidation state of iodine changes from

- 2 years ago

Answer is 7.75 *10^-4 mol L^-1 s^-1. But show me how to arrive at answer

- 2 years, 6 months ago

Since the reaction is of first order $$R=k[N_{2}O_{5}]$$

$$R=6.24\times 10^{-4} \times 1.25$$

$$R=7.75\times 10^{-4}~mol~ l^{-1} s^{-1}$$

- 2 years, 6 months ago

- 2 years, 6 months ago

Ya! Thank you very much.

- 2 years, 6 months ago