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The intial concentration of "N"_(2)"O"_(5) in the first order reaction, "N"_(2)"O"_(5)(g)to2"NO"_(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 :`k=(2.303)/(t)log""([A]_(0))/([A])=(2.303)/(t)log(["N"_(2)"O"_(5)]_(0))/(["N"_(2)"O"_(5)]_(t))=(2.303)/(60" min")log""(1.24xx10^(-2)"mol L"^(-1))/(0.2xx10^(-2)"mol L"^(-1))` `=(2.303)/(60)log6.2min^(-1)=(2.303)/(60)xx0.7924min^(-1)=0.0304min^(-1)` |
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