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:
- $k=1.5 \times 10^{-6} s^{-1}$
$k t=\ln \frac{100}{100-x} $
$\Rightarrow \quad \ln \frac{100}{100-x}=1.5 \times 10^{-6} s^{-1} \times 10 \times 60 \times 60 s=0.0054$
$\Rightarrow \quad \frac{100}{100-x}=1.055$
$\Rightarrow \quad x=5.25 \%$ reactant is converted into product.
Half-life $=\frac{\ln 2}{k}=\frac{0.693}{1.5 \times 10^{-6}}=462000 s=128.33 $ $h$
BETA
