JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
At \(345\, K\), the half life for the decomposition of a sample of a gaseous compound initially at \(55.5\, kPa\) was \(340\, s\). When the pressure was \(27.8 \,kPa\), the half life was fund to be \(170\, s\). The order of the reaction is \(......\) [integer answer]
- A \(1\)
- B \(0\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(0\)
Step-by-step Solution
Detailed explanation
\(t_{1 / 2} \times \frac{1}{\left[P_{0}\right]^{n-1}}\) \(\frac{t_{1}}{t_{2}}=\frac{\left(P_{2}\right)^{n-1}}{\left(P_{1}\right)^{n-1}}\) \(\frac{340}{170}=\left(\frac{27.8}{55.5}\right)^{ n -1}\) \(\Rightarrow 2=\frac{1}{(2)^{ n -1}}\) \(n =0\)
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