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For a reaction A to Products, starting with initial concentrations of 5xx10^(-3) M and25xx10^(-4)M, half-lives are found tobe 1.0 and 8.0 hour respectively. If we start with an initial concentration of 1.25xx10^(-3) M,what will be the half-lifeof the reaction ? |
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Answer» SOLUTION :For a REACTION of nth order, `t_(1//2)prop(1)/([A_(0)]^(n-1))` `:.((t_(1//2))_(1))/((t_(1//2))_(2))=([A_(0)]_(2)^(n-1))/([A_(0)]_(1)^(n-1))={([A_(0)]_(2))/([A_(0)]_(1))}^(n-1)` `(1)/(8)=((25xx10^(-4))/(5xx10^(-3)))^(n-1)=((1)/(2))^(n-1)" or "((1)/(2))^(3)=((1)/(2))^(n-1)" or "n-1=3" or "n=4` APPLYING the above formula again forinitial concentration `1.25xx10^(-3)"M and "5xx10^(-3)" M",` `((t_(1//2))_(1))/(1)=((5xx10^(-3))/(1.25xx10^(-3)))^(4-1)=(4)^(3)=64" hours, "i.e.,t_(1//2)=64" hours"` |
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