Q.

For the reaction, O2(g)+2F2(g)2OF2(g),Kp=4.1

If PO2(g)=0.116 atm and PF2(g)=0.0461 atm at equilibrium, what is the pressure of OF2(g)?

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a

0.101 atm

b

0.032 attn

c

0. 760 atm

d

166 atm

answer is B.

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Detailed Solution

The pressure on OF2 can be calculated as;4.1=POF220.116×(0.0461)2 POF2=0.032atm

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