Consider the following reaction at 150.0 degree C: I_2(g) + Br_2(g) 2IBr_(g) K_c

Question

Consider the following reaction at 150.0 degree C: I_2(g) + Br_2(g) 2IBr_(g) K_c = 280. Suppose that 0.500 mol IBr in a 1.00 L flask is allowed to reach equilibrium. What are the equilibrium concentrations of IBr, I_2, and Br_2? Methanol, CH_3OH, can be made by the reaction of CO with H_2(g): CO_(g) + 2 H_2(g) CH_3OH_(g) a. use thermochemical data in your textbooks appendices to calculate Delta H for this reaction. b. In order to maximize the equilibrium yield of methanol, would you use high or low temperature? c. Assuming equal pressures of co and H_2, how would the conversion of the gas mixture to methanol vary with total pressure?

Explanation / Answer

4) There is a 0.5 mol of IBr INA 1.0 lit flask so it's molarity is= 0.5 M (mol/lit) IBr.

-now write ICE table with initial concentration of all gaseous substances,

0.5-2X

Here X= unknown concentration of I2 and Br2.

The equilibrium constant for the given reaction is=280

Kc=[IBr]^2/[I2][Br2] .........1)

So, put all ICE table value's in equation 1to find X.

280=[0.5-2X]^2/[X*X]

280=[0.5-2X]^2/X^2

280*X^2= (0.5-2X)^2

280X^2= 0.5^2 - 2(2X)(0.5) + 4X^2

276X^2+ 2X-0.25 =0

So,X=0.0267

put this value in ICE table,

I2=0.0267M

Br2=0.0267M

IBr=0.5-(2*0.0267)=0.5-0.0534 =0.4466M

I2 Br2 2IBr I 0 0 0.5 C +X +X -2X E X X

0.5-2X

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