If 0.0600 mol of IBr is placed in a 1.0-L container, the partial pressure of I₂(g) after equilibrium is reached is 0.176 atm.
0.0600 mol of IBr is placed in a 1.0-L container. The initial molar concentration is:
[tex][IBr]i = \frac{0.0600mol}{1.0L} = 0.060M[/tex]
We will make an ICE chart for the following reaction.
2 IBr(g) ⇌ I₂(g) + Br₂(g)
I 0.060 0 0
C -2x +x +x
E 0.060-2x x x
The concentration equilibrium constant (Kc) is:
[tex]Kc = 8.50 \times 10^{-3} = \frac{[I_2][Br_2]}{[IBr]^{2} } = \frac{x^{2} }{(0.060-x)^{2} } \\\\x = 0.00507[/tex]
The concentration (C) of I₂ at equilibrium is x = 0.00507 M. We can calculate its partial pressure (P) using the following expression.
[tex]P = C \times R \times T = 0.00507mol/L \times \frac{0.08206atm.L}{mol.K} \times 423 K = 0.176 atm[/tex]
where,
If 0.0600 mol of IBr is placed in a 1.0-L container, the partial pressure of I₂(g) after equilibrium is reached is 0.176 atm.
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