Chemical Kinetics - Result Question 77
####76. A first order gas reaction has $k=1.5 \times 10^{-6}$ per second at $200^{\circ} C$. If the reaction is allowed to run for $10 h$, what percentage of the initial concentration would have change in the product? What is the half-life of this reaction?
$(1987,5 M)$
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Solution:
- $k=1.5 \times 10^{-6} s^{-1}$
$$ \begin{gathered} k t=\ln \frac{100}{100-x} \ \Rightarrow \quad \ln \frac{100}{100-x}=1.5 \times 10^{-6} s^{-1} \times 10 \times 60 \times 60 s=0.0054 \end{gathered} $$
$\Rightarrow \quad \frac{100}{100-x}=1.055$
$\Rightarrow \quad x=5.25 %$ reactant is converted into product.
Half-life $=\frac{\ln 2}{k}=\frac{0.693}{1.5 \times 10^{-6}}=462000 s=128.33 h$
