A chemist introduced 0.027 mol of H 2 , 0.046 mol of I 2 , and 0.55 mol of HI in

Question

A chemist introduced 0.027 mol of H2, 0.046 mol of I2, and 0.55 mol of HI into a 2.0 L container and heated it to 783 K. At this temperature, Kc = 46.0 for the reaction H2(g) + I2 (g) ? 2HI (g)

Which all apply? please show work

1.the reaction is already at equilibrium.

2.more HI will be formed before equilibrium is reached.

3.more HI will be decomposed before equilibrium is reached.

4.the reaction must proceed to the right to reach equilibrium.

5.the reaction must proceed to the left to reach equilibrium.

6.more H2 will be consumed before equilibrium is reached.

7.more H2 will be formed before equilibrium is reached.

Explanation / Answer

Reaction:

H2 + I2 <----> 2HI

initially 0.0135 0.023 0.275

finally 0.0135 -x 0.023- x 0.275+2 x

let Q=[HI]/[ H2]*[ I2] at initial condition

Q=855.668

now Q>ka

therefore concentration of HI is greater than its equilibrium concentration and reaction will proceed in left direction.

hence correct options are 3,5,&7

ka = [HI]/[ H2]*[ I2]= 46= (0.0275+2x)/(0.0135-x)*(0.023-x)

x=0.125 & -0.045

x=-0.045 (as reaction proceeds to left hand side)

at equilibrium

[HI]=0.185M

[ H2] = 0.0585M

[ I2] =0.068M

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