Consider the reaction, which takes place at a certain elevated temperature co(g)+nh3(g)⇌hconh2(g), kc=0.780co(g)+nh3(g)⇌hconh2(g), kc=0.780 if a reaction vessel initially contains only coco and nh3nh3 at concentrations of 1.00 mm and 2.00 mm, respectively, what will the concentration of hconh2hconh2 be at equilibrium?

The concentration of HCONH2 at the equilibrium will be 0.5335 M

Explanation:

Step 1: Data given

Kc = 0.780

The reaction vessel initially contains only CO and NH3

[CO] = 1.00 M

[NH3]= 2.00 M

Step 2: The balanced equation

CO(g) + NH3(g) ⇌ HCONH2(g),

Step 3: The initial concentrations

[CO] = 1.0 M

[NH3] = 2.0 M

[HCONH2] = 0 M

Step 4: Calculate the concentration at the equilibrium

For 1 mol CO we need 1 mol 1 mol NH3 to produce 1 mol HCONH2

[CO] = 1.0 -X M

[NH3] = 2.0 - X M

[HCONH2] = X M

Step 5: Calculate the concentration of HCONH2

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

0.780 = X / ((1.0 - X)(2.0- X))

X = 0.5335

[CO] = 1.0 -0.5335 = 0.4665  M

[NH3] = 2.0 - 0.5335 = 1.4665 M

[HCONH2] = 0.5335 M

0.5335 / 0.4665 * 1.4665 = 0.780

The concentration of HCONH2 at the equilibrium will be 0.5335 M

answer; i believe that the correct answer is (a) potassium(k)