1.

For the reaction 2 N_(2) O_(5(g)) to 4 NO_(2 (g)) to O_(2(g)) , if concentration of NO_(2) in 100 seconds is increased by 5.2 xx 10^(-3) m . Then the rate of reaction will be

Answer»

`1.3 xx 10^(-5) ms^(-1)`
`5 xx 10^(-4) ms^(-1)`
`7.6 xx 10^(-4) ms^(-1)`
`2 xx 10^(-3) ms^(-1)`

Solution :`2N_(2)O_(5(G)) to 4NO_(2 (g)) + O_(2(g))`
Rate of REACTION with RESPECT to `NO_(2)`
`=(1)/(4) (d[NO_(2)])/(DT) = (1)/(4) xx (5.2 xx 10^(-3))/(100) = 1.3 xx 10^(-5) ms^(-1)`.


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