Formamide decomposes at high temperature. If 0.186 mol of formamide (HCONH2) dissociates in a 2.16 L flask at 400 K, what are the concentrations of all species present at equilibrium at 400 K? (hint: calculate concentrations first) (b) What is the total pressure in the container at equilibrium?

a) [COHNH₂] = 0.001 mol/L, [NH₃] = [CO] = 0.085 mol/L

b) 5.59 atm

Explanation:

a) The decomposition reaction of formamide is the following:

COHNH₂(d)  ⇆  NH₃(g) + CO(g)    

The equilibrium constant of the reaction above is:

The initial concentration of formamide is:

Where: η is the number of moles and V is the volume

Now, in the equilibrium the concentration of all species is:

COHNH₂(d)  ⇆  NH₃(g) + CO(g)  

0.086 - x              x             x

By solving the above equation for x we have:

x = 0.085 mol/L = [NH₃] = [CO]

[COHNH₂] = 0.086 - 0.085 = 0.001 mol/L

Therefore, the concentrations of all species present at equilibrium at 400 K is [NH₃] = [CO] = 0.085 mol/L and [COHNH₂] = 0.001 mol/L.

b) To find the total pressure in the container we need to find first the constant Kp as follows:

Where R is the gas constant = 0.082 Latm/(Kmol), T is the temperature = 400 K and Δn = 1

Now, the total pressure is:

The pressure of COHNH₂ can be found using Ideal Gas Law:

Using the equilibrium constant we can find the pressure of NH₃ and CO:

COHNH₂(d)  ⇆  NH₃(g) + CO(g)  

2.82 - x              x             x

By solving the above equation for x we have:

Thus, the total pressure is:

Hence, the total pressure in the container at equilibrium is 5.59 atm.

I hope it helps you!

it would be d. henry moseley

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