Question

for a zero order reaction and first order reaction half life are in ratio of 4:1 .calculate ratio of time taken to complete 87.5 % reaction for zero order:first order​

Answer 1

Answer:

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Answer 2

Answer:

The ratio of time taken to complete 87.5 % reaction for zero order: first order​ is 7:3

Explanation:

Given

$(t_{1/2)_{0} /(t_{1/2}) _{1} =4/1$

time- taken by zero-order reaction for decay=(t₁/₂)₀ let

time- taken by first-order reaction for decay=(t₁/₂)₁ let
Rate of decay of both=87.5%

To find the ratio of time taken to complete 87.5 % reaction for zero order: first order​ (t₁/₂)₀/(t₁/₂)₁

Solution:

If,  rate constant for a zero-order reaction(k)=k₀

and, rate constant for first order reaction=k₁

The half-life for zero-order is,

$(t_{1/2})_{0} = \frac{C_{0} }{2k_{0} }$

and the half-life of first-order is,

$(t_{1/2})_{1} =\frac{ln^{2} }{k_{1} }$

Then,

According to the integrated rate law for zero-order,

Ct=C₀ - k₀t₀----eq.1  

where Ct is the final concentration of reactant and C₀ is the initial           concentration of reactant

% of the remaining concentration of reactant=100-87.5=12.5% of C₀

Ct=12.5% of C₀

From eq.1

Ct=C₀ - k₀t₀

12.5/100C₀=C₀ - k₀t₀

C₀ - (12.5/100C₀)=k₀t₀

87.5C₀/100=k₀t₀

t₀=C₀/k₀(87.5/100)

Again,

Integrated law for first-order

Ct=C₀e^-kt---eq. 2

12.5C₀/100=C₀e^-kt₁

12.5/100=e^-kt₁

ln (12.5/100)=ln e^-kt₁

ln12.5/100=-kt₁  [∵ln e^x=x]

ln8=kt₁   [ln1/m=-lnm]

t₁=1/k ln8

Now,

t₀/t₁=t₀=C₀/k₀(87.5/100)/1/k ln8

⇒ C₀/2k₀ ₓ 87.5/100/ln2/k ₓ 3

⇒ 4/1×87.5/50×3

⇒70/30

⇒7/3

⇒7:3

Thus, the ratio of time taken to complete 87.5 % reaction for zero order: first order​ is 7:3.