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In a 7.0 Levacuated chamber, 0.50 mol H(...

In a 7.0 Levacuated chamber, 0.50 mol `H_(2)` and 0.50 mol `I_(2)` react at `427^(@)` C.
`H_(2) (g) + I_(2) (g) hArr HI (g) ` + Heat , At the given temperature, `K_(c)` = 49 for the reaction.
What is the total pressure (atm) in the chamber ?

A

83.14

B

831.4

C

8.21

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
PV=nRT
`implies p=(nRT)/(V)=(1 xx 0.0821 xx 700)/(7)=8.21` atm
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