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Chemistry Question on Chemical Kinetics

For the reaction {A_{\left(g\right)}−> -> B_{\left(g\right)}++C_{\left(g\right)}, } the rate constant is given as (P is initial pressure and P is pressure at time t)

A

k=2.303tk=\frac{2.303}{t} log PiPt\frac{P_{i}}{P_{t}}

B

k=2.303tk=\frac{2.303}{t} log Pi(2Pi−Pt)\frac{P_{i}}{\left(2P_{i}-P_{t}\right)}

C

k=2.303tk=\frac{2.303}{t} log 2Pi−PtPi\frac{2P_{i}-P_{t}}{P_{i}}

D

k=2.303tk=\frac{2.303}{t} log Pi−Pt2Pi\frac{P_{i}-P_{t}}{2P_{i}}

Answer

k=2.303tk=\frac{2.303}{t} log Pi(2Pi−Pt)\frac{P_{i}}{\left(2P_{i}-P_{t}\right)}

Explanation

Solution

For the reaction, A(g)−>B(g)+C(g) {A_{(g)}->B_{(g)} + C_{(g)}} t=0Pi00 t(Pi−x)xx\begin{matrix}t=0&P_{i}&0&0\\\ t&\left(P_{i}-x\right)&x&x\end{matrix} Given: P is initial pressure at time, t = 0 and P is pressure at time ???? P = (P - x) + x + x = P + x x = P - P Here, P = P - x ?? P - (P - P) \quadP = 2P - P k=2.303tk=\frac{2.303}{t} log PiPA\frac{P_{i}}{P_{A}} k=2.303tk=\frac{2.303}{t} log Pi(2Pi−Pt)\frac{P_{i}}{\left(2P_{i}-P_{t}\right)}