Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

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 :

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

Read more about Chemical Reaction

https://brainly.com/question/11231920

#SPJ1

We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.