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Suppose that 0.25M IBr in a flask is allowed to reach equilibrium at 150oC. The equilibrium concentration of Br2 is 0.013 M. Determine the Kc at this temperature. 2IBr(g) ←→ I₂(g) + Br₂(g)

Sagot :

okpalawalter8

Suppose that 0.25M IBr in a flask,  the Kc at the given temperature is mathematically given as

Kc=0.00336814413

What is the Kc at the given temperature?

Generally, the equation for Chemical Reaction is mathematically given as

2IBr--->T2+8r_2

Therefore

K={I2*Br2}/IBR^2

K=x^2/(0.25-2x)^2

Kc=0.13M^2/0.224M^2

Kc=0.00336814413

In conclusion,  the Kc at the given temperature

Kc=0.00336814413

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