    Question

# A(g)→2B(g)+C(g) Initially at t = 0 gas A was present along with some amount of gas C. At At t = 0, the mole fraction of gas C is 13. After some time t=t1, the total pressure is half of the final total pressure at t=tx (a very long time). Assume this decomposition to follow first order kinetics (at a constant temperature). It is also given at t=tx, the final total pressure is 35 bar. Rate constant (k)=(log64−log49)s−1. Value of t1 in seconds is:

A
2.15 s
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B
1.5 s
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C
2.3 s
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D
1.15 s
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Solution

## The correct option is D 1.15 sA(g)→2B(g)+C(g) 23Pi 0 13Pi at t=0 23Pi−p 2p 13Pi+p at t1 23Pi−px 2px 13Pi+px at tx from given data pi+2p=352 pi+2px=35 after so long time 23Pi−px=0 by solving we get Pi=15 2p=2.5 We know that for a first order reaction at time t, k=1tln[intial][final] log64−log49=1tln[23pi][23pi−p] t=1.15  Suggest Corrections  0      Related Videos   Rate Constant
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