The equilibrium pressure of NH₄CN_(s)==NH_3(g)+HCN_(g) is 4atm.In an experiment if NH₄CN_(s) is allowed to decompose in presence of NH_3(g) at 3atm,then what is the pressure of HCN at new and old equilibrium states?

To tackle this question, we need to analyze the equilibrium reaction of ammonium cyanide (NH4CN) decomposing into ammonia (NH3) and hydrogen cyanide (HCN). The equilibrium expression and the given conditions are key to finding the new pressure of HCN after the decomposition in the presence of NH3 at 3 atm.

The Reaction and Initial Conditions

The decomposition of NH4CN can be represented as:

NH4CN(s) ⇌ NH3(g) + HCN(g)

According to the problem, the equilibrium pressure of NH3 and HCN in the original state is 4 atm. This means that at equilibrium:

• P(NH3) = x atm
• P(HCN) = y atm

From the equilibrium condition, we can establish that:

x + y = 4 atm

New Experiment Conditions

In the experiment, NH4CN(s) is allowed to decompose in the presence of NH3 at a pressure of 3 atm. This means that the initial pressure of NH3 is already 3 atm, which will affect the system's equilibrium.

At the onset of the reaction, we can denote the pressure of HCN produced as 'y'. Thus, we have:

• P(NH3) = 3 atm (initially present)
• P(HCN) = y atm (produced from the decomposition)

The new total pressure must still satisfy the equilibrium condition. Therefore, the new equilibrium condition can be expressed as:

3 atm + y = 4 atm

Solving for HCN Pressure

To find the pressure of HCN at the new equilibrium state, we can rearrange the equation:

y = 4 atm - 3 atm = 1 atm

Summary of Equilibrium States

At the original equilibrium state:

• P(NH3) = 4 atm (when both gases are in equilibrium)
• P(HCN) = 0 atm (as it wasn't produced yet)

At the new equilibrium state (after accounting for the decomposition in the presence of NH3):

• P(NH3) = 3 atm (remains constant due to external addition)
• P(HCN) = 1 atm (produced from the decomposition)

Thus, to summarize, at the new equilibrium state after allowing NH4CN to decompose in the presence of NH3 at 3 atm, the pressure of HCN will be 1 atm. This demonstrates how equilibrium shifts when the conditions change, and how the principles of gas laws and equilibrium can be applied to solve for unknown pressures in a chemical reaction.