1. Tardigrade
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  3. Chemistry
  4. The rate constant for the first order decomposition of a certain reaction is described by the equation: text log textk (( texts)-1) = text14.34 - ( text1.25 × ( text10)4 text K/ textT) At what temperature will its half-life period be text256 min ?

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Q. The rate constant for the first order decomposition of a certain reaction is described by the equation:

$\text{ log } \text{k } \left(\left(\text{s}\right)^{-1}\right) = \text{14.34} - \frac{\text{1.25} \times \left(\text{10}\right)^{4} \text{ K}}{\text{T}}$

At what temperature will its half-life period be $\text{256 min}$ ?

NTA AbhyasNTA Abhyas 2020Chemical Kinetics

Solution:

When half-life $=256 min$

$\begin{aligned} k &=\frac{\ln 2}{ t _{1 / 2}}=\frac{0.693}{256 \times 60} s ^{-1} \\ &=4.5 \times 10^{-5} s ^{-1} \\ \Rightarrow & \frac{1.25 \times 10^{4}}{T}=14.34-\left[\log 4.5 \times 10^{-5}\right] \\ \Rightarrow & 14.34-[-5+0.65] \\ \Rightarrow & 19 \cdot 34-0 \cdot 65=18 \cdot 69 \\ T =& \frac{12500}{18.69}=669 K \end{aligned}$