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Question

# At 800∘C, the following equilibrium is established as F2(g)⇌2F(g) The composition of equilibrium may be determined by measuring the rate of effusion of the mixture through a pin hole. It is found that at 800∘C and 1 atm mixture effuses 1.6 times as fast as SO2 effuses under the similar conditions. (Atomic weight of F =19 amu). What is the value of Kp?

A
0.315 atm
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B
0.685 atm
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C
0.460 atm
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D
1.490 atm
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Solution

## The correct option is D 1.490 atmrmixrSO2 = (MSO2Mmix)12 2.56 = 64Mmix Mmix = 25 Let's say mole fraction of F2 is x 25 = 38× x + (1 − x)×191 x=0.315 Kp = (PF)2PF2 = (0.685)20.315 = 1.49 atm

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