Consider the reaction co(g)+nh3(g)âhconh2(g), kc=0.900 if a reaction vessel initially contains only co and nh3 at concentrations of 1.00 m and 2.00 m, respectively, what will the concentration of hconh2 be at equilibrium? express your answer with the appropriate units.

See explanation below

Explanation:

Let's write the equilibrium reaction:

CO(g) + NH3(g) <> HCONH2(g)   Kc = 0.9

We know the initial concentrations which are 1 and 2 M. In order to know the concentration of equilibrium, we need to do an ICE chart, and then, write the equilibrium expression for this reaction to calculate the concentration.

The ICE chart:

       CO(g) + NH3(g) <> HCONH2(g)   Kc = 0.9

i)        1              2                         0

c)       -x           -x                          x

e)      1-x          2-x                        x

The equilibrium expression:

Kc = [HCONH2] / [CO][NH3]

Replacing the values we have:

0.9 = x / (1-x)(2-x)

0.9(1-x)(2-x) = x

0.9(x²-3x+2) = x

0.9x² - 2.7x + 1.8 = x

0.9x² - 3.7x + 1.8 = 0

From here, we solve for x using the general formula for a quadratic expression:

x = -b ±√b² - 4ac / 2a

Replacing the values here:

x = 3.7 ±√(3.7)² - 4 * 0.9 * 1.8 / 2*0.9

x = 3.7 ±√13.69 - 6.48 / 1.8

x = 3.7 ± 2.69 / 1.8

x1 = 3.7 - 2.69 / 1.8 = 0.56

x2 = 3.7 + 2.69 / 1.8 = 3.55

As 3.55 is > 1 and 2 from the initial concentrations, the correct value is x1 therefore:

[HCONH2] = 0.56 M

c

explanation: