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The following reaction was carried out a...

The following reaction was carried out at `44^(@)C` :
`N_(2)O_(5)rarr2NO_(2)+_(1//2)O_(2)`
The concentration of `NO_(2)` is `6.0xx10^(-3)` M afte minutes of the start of reaction . Calculate the rat of production of `NO_(2)` over the first ten minutes of the reaction

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The correct Answer is:
`6.0xx10^(-4) "mol" L^(-1) "min" ^(-1)`
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For the reaction : 2N_(2)O_(5)rarr4NO_(g)+O_(2)(g) if the concentration of NO_(2) increases by 5.2xx10^(-3)M in 100 sec, then the rate of reaction is :

For the reactio 2N_(2)O_(5)to4NO_(2)+O_(2) The concentration of N_(2)O_(5) changes from 3.0 M to 2.75 M in 30 minute. Calculate rate of formation of NO_(2)

The reaction, 2N_(2)O_(5)rarr 4NO_(2)+O_(2) , shows an increase in concentration of NO_(2) by 20xx10^(-3) mol litre^(-1) in 5 second. Calculate: (a) rate of appearance of NO_(2) , (b) rate of reaction and (c ) rate dissappearance of N_(2)O_(5) .

For the reaction 2N_(2)O_(5)(g) to 4NO_(2) (g) + O_(2)(g) , the rate of formation of NO_(2)(g) is 2.8 xx 10^(-3) Ms^(-1) . Calculate the rate of disapperance of N_(2)O_(5)(g) .

For the reaction N_(2)O_(5) rarr 2NO_(2) + (1)/(2) O_(2) , the rate of disappearance of N_(2)O_(5) is 6.25 xx 10^(-3) "mol L"^(-1) s^(-1) . The rate of formation of NO_(2) and O_(2) will be respectively.

Using the concentration time equation for a first order reaction : The decomposition of N_(2)O_(5) to NO_(2) and O_(2) is first order, with a rate constant of 4.80xx10^(-4)//s att 45^(@)C N_(2)O_(5)(g)rarr2NO_(2)(g)+(1)/(2)O_(2)(g) (a) If the initial concentration of N_(2)O_(5) is 1.65xx10^(-2)mol//L , what is its concentration after 825 s ? (b) How long would it take for the concentration of N_(2)O_(5) to decrease to 100xx10^(-2) mol L^(-1) from its initiqal value, given in (a) ? Strategy : Since this reaction has a first order rate law, d[N_(2)O_(5)]//dt = k[N_(2)O_(5)] , we can use the corresaponding concentration time equation for a first order reaction : k = (2.303)/(t) log ([N_(2)O_(5)]_(0))/([N_(2)O_(5)]_(t)) In each part, we substitute the know quantities into this equation and solve for the unkbnown.

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