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If the half-life for a reaction is 20. Seconds, what would be the second half-life, assuming the reaction is either zero, first, or second-order?

Short Answer

Expert verified

The second half-life for the zero, first, and second-order reactions is 10, 40 and 40 s, respectively.

Step by step solution

01

Calculating the second half-life for the first-order reaction

The first half-life t12is 20s.

The second half-life is the time needed for the conversion from 50% to 25%.

For the second half-life, A=A0/4

For zero-order,

t12=A02kt2ndhalflife=A022k

Second half-life

t2ndhalflife=202t2ndhalflife=10s

The second half-life for the zero-order reaction is the 10 seconds.

02

Calculating the second half-life for a first-order reaction

For the first order,

t12=0.693kk=0.693t12

k=0.69320k=0.03465

Second half-life,

t2ndhalflife=1.386kt2ndhalflife=1.3860.03465t2ndhalflife=40s

The second half-life for the first-order reaction is 40 seconds.

03

Calculating the second half-life for a second-order reaction

For second-order,

t12=1kA0

Second half-life

role="math" localid="1663761517108" t2ndhalflife=2kA0=21kA0=2t12=220=40s

The second half-life for the second-order reaction is 40 seconds.

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Most popular questions from this chapter

Hydrogen peroxide and the iodide ion react in acidic solution as follows:

H2O2aq+3I-aq+2H+aqI3-aq+2H2Ol

The kinetics of this reaction were studied by following the decay of the concentration of and the constructing plots of lnH2O2versus time. All the plots were linear and all solutions had H2O2=8.0×10-4mol/L . The slopes of these straight lines depended on the initial concentrations of H+ and I- . The result follow:

The rate law for this reaction has the form

Rate=- dH2O2dt=K1+K2H+I-mH2O2n

  1. Specify the orders of this reaction with respect toH2O2 and I-.
  2. Calculate the values of the rate constantsK1 and K2.
  3. What reasons could there be for the two-term dependence of the rate onH+ ?

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The reaction H2SeO3(aq)+6I-(aq)+4H+(aq)Se(s)+2I3-(aq)+3H2O(l)was studied at 0°C, and the following data were obtained:

[H2SeO]0(mol/L)
[H+]0(mol/L)
[I-](mol/L)
InitialRate(molL-1s-1)
localid="1663839197909" 1.0×10-4
localid="1663839994858" 2.0×10-2
localid="1663840803384" 2.0×10-2
role="math" localid="1663841146761" 1.66×10-7
2.0×10-4
localid="1663840000906" 2.0×10-2
2.0×10-2
3.33×10-7
3.0×10-4
localid="1663840007628" 2.0×10-2
2.0×10-2
4.99×10-7
1.0×10-4
localid="1663840013722" 4.0×10-2
2.0×10-2
6.66×10-7
1.0×10-4
localid="1663840021033" 1.0×10-2
2.0×10-2
0.42×10-7
1.0×10-4
localid="1663840026824" 2.0×10-2
4.0×10-2
13.2×10-7
1.0×10-4
localid="1663840033291" 1.0×10-2
4.0×10-2
3.36×10-7

The relationships hold only if there is an insignificant amount of I3- present. What is the rate law and the value of the rate constant?

(Assumethatrate=-d[H2SeO3]dt.)

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Use LE model to describe the bonding in hydrogen cyanide and phosgene.

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