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The initial concentration of N_(2)O_(5) in the following first order reaction N_(2)O_(5(g))to2NO_(2(g))+(1)/(2)O_(2(g)) was 1.24xx10^(-2) Mol L^(-1) at 318 K.The concentration of N_(2)O_(5) after 60 minutes was 0.20xx10^(-2) mol L^(-1). calculate the rate constant of the reaction at 318 K. |
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Answer» Solution :Given reaction is first ORDER in it k=(?) INITIAL t=0 time ,the concentration of `N_(2)O_(5)` is `1.24xx10^(-2) mol L^(-1)` So.`t_(1)=0` and `[R]_(1)=1.24xx10^(-4)` mol `L^(-1)` After 60 MINUTES the concentration of `N_(2)O_(5)=0.2xx10^(-2) mol L^(1)` So, `t_(2)=60` minutes and `[R]_(2)=0.2xx10^(-2)` mol `L^(-1)` For a first order reaction , `log([R]_(1))/([R]_(2))=(k(t_(2)-t_(2)))/(2.303)` `therefore log (1.24xx10^(-2))/(0.2xx10^(-1))=k((60-0)"MINUTE")/(2.303)` `therefore log 6.2=k(60 "minute")/(2.303)` `therefore 0.7924=k(60)/(2.303)` `therefore k=(0.7924xx2.303)/(60 "minute") =0.0304 "minute"^(-1)` |
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