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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?

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carolpire

Answer:

Concentration of HCONH2 will be 0.5333 MM

Explanation:

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Answer:

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

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