1.

Decomposition of H_(2)O_(2) follows a first oeder reaction.In fifty minutes the concentration of H_(2)O_(2) decreases from 0.5 to 0.125 M in one such decomposition .When the concentration of H_(2)O_(2) reaches 0.05 M ,the rate of formation of O_(2) will be:

Answer»

`1.34xx10^(-2)` mol `min^(1)`
`6.93xx10^(-2)`mol `min^(-1)`
`6.93xx10^(-4)` mol `min^(-1)`
2.66 L `min^(-1)` at STP

Solution :Here solution is given assuming UNIT of rate is the option to be `mol^(-1)L^(-1) min^(-1)`
For first order reaction K=`(1)/(50)` in `(0.5)/(0.125)`
K=`(In4)/(50) min^(-1)`
Now R=`K[H_(2)O_(2)]^(1)` `=2xx0.693xx10^(-3) mol L^(-1)min^(-1)`
and `H_(2)O_(2)toH_(2)O+(1)/(2)O_(2)`
So rate of reaction.
`R=(R_(O_(2))("Rate formation of"O_(2)))/(((1)/(2)))`
or `R_(O_(2))=6.93xx10^(-4)mol L^(-1)min^(-1)`


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