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Question

Consider the following initial rate data for the decomposition of compound AB to give A and B:

AB]0(mol/L) Initial Rate (mol L1s1)\lfloor \mathrm { AB } ] _ { 0 } ( \mathrm { mol } / \mathrm { L } ) \quad \quad \text { Initial Rate (mol } \mathrm { L } ^ { - 1 } \mathrm { s } ^ { - 1 } )

0.2003.20×1030.4001.28×1020.6002.88×102\begin{array} { l l } { 0.200 } &&&&&& { 3.20 \times 10 ^ { - 3 } } \\ { 0.400 } &&&&&& { 1.28 \times 10 ^ { - 2 } } \\ { 0.600 } &&&&&& { 2.88 \times 10 ^ { - 2 } } \end{array}

Determine the half-life for the decomposition reaction initially having 1.00 M AB present.

Solution

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Answered 2 years ago
Answered 2 years ago
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rate2rate1=[AB]x[AB]x=(0.4000.200)x\dfrac{\text{rate}_2}{\text{rate}_1} =\dfrac{\left[\text{AB}\right]^{\text{x}}}{\left[\text{AB}\right]^{\text{x}}} = \left(\dfrac{0.400}{0.200}\right)^{\text{x}}

2x=1.28×1023.20×103=42^{\text{x}} = \dfrac{1.28 \times 10^{-2}}{3.20 \times 10^{-3}} = 4

x=2\boxed{\text{x} = 2}

Now that we know we have second-order reaction we can determine reaction rate constant:

rate=k[AB]2\text{rate} = \text{k}\left[\text{AB}\right]^2

k=rate[AB]2=3.20×103 mol L1 s1(0.200 mol/L)2\text{k} = \dfrac{\text{rate}}{\left[\text{AB}\right]^2} = \dfrac{3.20 \times 10^{-3} \ \text{mol} \ \text{L}^{-1} \ \text{s}^{-1}}{\left(0.200 \ \text{mol/L}\right)^2}

k=0.08 L mol1 s1\boxed{\text{k} = 0.08 \ \text{L} \ \text{mol}^{-1} \ \text{s}^{-1}}

Now we can determine half-life:

t1/2=1k[AB]0=10.08 L mol1 s1×1.00 M\text{t}_{1/2} = \dfrac{1}{\text{k}\left[\text{AB}\right]_0} = \dfrac{1}{0.08 \ \text{L} \ \text{mol}^{-1} \ \text{s}^{-1} \times 1.00 \ \text{M}}

t1/2=12.5 s\boxed{\text{t}_{1/2} = 12.5 \ \text{s}}

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