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Choose the Best Answer · Q21

Q.In a homogeneous reaction A→B+C+DA \rightarrow B+C+D, the initial pressure was P0P_0 and after time t it was P. The expression for rate constant in terms of P0P_0, P and t will be

(a) k=(2.303t)log⁡(2P03P0−P)k=\left(\dfrac{2.303}{t}\right)\log\left(\dfrac{2P_0}{3P_0-P}\right)
(b) k=(2.303t)log⁡(2P0P0−P)k=\left(\dfrac{2.303}{t}\right)\log\left(\dfrac{2P_0}{P_0-P}\right)
(c) k=(2.303t)log⁡(3P0−P2P0)k=\left(\dfrac{2.303}{t}\right)\log\left(\dfrac{3P_0-P}{2P_0}\right)
(d) k=(2.303t)log⁡(2P03P0−2P)k=\left(\dfrac{2.303}{t}\right)\log\left(\dfrac{2P_0}{3P_0-2P}\right)
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Step 1. Let x = the drop in A's own partial pressure. Then: partial pressure of A remaining =P0−x=P_0-x; B, C and D (each coefficient 1) each have partial pressure xx. Total pressure P=(P0−x)+x+x+x=P0+2xP=(P_0-x)+x+x+x=P_0+2x.

Step 2. Solving for x: x=P−P02x=\dfrac{P-P_0}{2}.

Step 3. Partial pressure of A remaining =P0−x=P0−P−P02=2P0−P+P02=3P0−P2=P_0-x=P_0-\dfrac{P-P_0}{2}=\dfrac{2P_0-P+P_0}{2}=\dfrac{3P_0-P}{2}. …

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