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?

#Chemistry #JEEchemistry

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Comments

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

Ashley Shamidha - 6 years, 2 months ago

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

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

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

Kyle Finch - 6 years, 2 months ago

Was it helpful????

Kyle Finch - 6 years, 2 months ago

Ya! Thank you very much.

Ashley Shamidha - 6 years, 2 months ago

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

Ashish Gaur - 5 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$