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Question: The reaction A(s) \(\rightarrow\) 2 B(g) + C(g) is first order. The pressure after 20 min. and after...

The reaction A(s) →\rightarrow 2 B(g) + C(g) is first order. The pressure after 20 min. and after very long time are 150 mm Hg and 225 mm Hg. The value of rate constant and pressure after 40 min. are :

A

0.05 In 1.5 min–1, 200 mm

B

0.5 ln 2 min–1, 300 mm

C

0.05 In 3 min–1, 300 mm

D

0.05 In 3 min–1 , 200 mm

Answer

0.05 In 3 min–1 , 200 mm

Explanation

Solution

A (s) →  2 B (g) + C (g) \text{A }(s)\ \overset{\quad\ \quad}{\rightarrow}\text{ 2 B }(g)\text{ + C }(g)\

K=1tIn(P∞−P0P∞−Pt)P0=0Pt=150P∞=225K = \frac{1}{t}In\left( \frac{P_{\infty} - P_{0}}{P_{\infty} - P_{t}} \right) P_{0} = 0 P_{t} = 150 P_{\infty} = 225

120In[225225−150]⇒120In(22575)=0.05In3\frac{1}{20}In\left\lbrack \frac{225}{225 - 150} \right\rbrack \Rightarrow \frac{1}{20}In\left( \frac{225}{75} \right) = 0.05In3

120In(22575)=120In(225225×P40)⇒(22575)2=225225×P40\frac{1}{20}In\left( \frac{225}{75} \right) = \frac{1}{20}In\left( \frac{225}{225 \times P_{40}} \right) \Rightarrow \left( \frac{225}{75} \right)^{2} = \frac{225}{225 \times P_{40}}

22575×75=1225−P40⇒P40=200\frac{225}{75 \times 75} = \frac{1}{225 - P_{40}} \Rightarrow P_{40} = 200