Question

Formamide, used in the manufacture of pharmaceuticals, dyes, and agricultural chemicals, decomposes at high temperatures. $$\begin{array}{r} \mathrm{HCONH}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \\ K_{\mathrm{c}}=4.84 \text { at } 400 \mathrm{K} \end{array}$$ If \(0.186 \mathrm{mol} \mathrm{HCONH}_{2}(\mathrm{g})\) dissociates in a 2.16 Lflask at 400 K, what will be the total pressure at equilibrium?

Short answer

The total pressure at equilibrium under given conditions is approximately 5.61 atm

Step-by-step solution

Understanding the information

Formamide (HCONH2) decomposes into NH3 and CO. It is given that initially, 0.186 mol of HCONH2 was present and it dissociates at 400 K in a 2.16 L flask. At equilibrium, the equilibrium constant (Kc) is 4.84.

Set up the equilibrium table for equilibrium concentrations

At start, the moles of HCONH2 equals 0.186 mol. NH3 and CO are zero. At equilibrium, the amount of HCONH2 that dissociates can be represented as x. So, at equilibrium the amounts of gases will be: HCONH2, NH3 and CO will be 0.186 - x, x and x mol respectively.

Substitute given quantities into equilibrium constant expression

The expression for Kc is [NH3][CO]/[HCONH2]. Substituting values we get: 4.84=(x/2.16)*(x/2.16)/((0.186-x)/2.16). Simplifying this equation will allow us to solve for x.

Solve for x

Solving the equation, we get x approximately equal to 0.115 mol. This means, at equilibrium, we have 0.115 mol of NH3 and CO each and 0.071 mol of HCONH2.

Calculate total pressure at equilibrium

Total pressure can be calculated using the Ideal Gas Law: P = nRT/V. In this case, n is total number of moles of all gases, R is the gas constant (0.0821 L·atm/K·mol), T is the temperature in Kelvin and V is the volume in liters. Substituting for n = 0.115 (NH3) + 0.115 (CO) + 0.071 (HCONH2) = 0.301 mol, T = 400 K, and V = 2.16 L, the equation yields the total pressure at equilibrium.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle used to relate the pressure, volume, temperature, and amount of a gas. The equation is given by:
\[ PV = nRT \]Here,

• \( P \) is the pressure
• \( V \) is the volume
• \( n \) is the number of moles of the gas
• \( R \) is the ideal gas constant, which is 0.0821 L·atm/K·mol
• \( T \) is the temperature in Kelvin

This law simplifies calculations by allowing you to find one of these variables if the others are known.
In the context of chemical equilibrium, it helps calculate the total pressure exerted by gases. It is essential to ensure that the temperature is always in Kelvin and the volume is consistent with the units of the gas constant. This makes the law a powerful tool in solving equilibrium problems that deal with gases.

Equilibrium Constant

The equilibrium constant, represented as \( K_c \), is a value that describes the ratio of concentrations of products to reactants at equilibrium for a reversible chemical reaction. The equation for the equilibrium constant is:
\[ K_c = \frac{[products]}{[reactants]} \]The brackets indicate molar concentration. This constant only changes with temperature. For the decomposition of formamide:
\[ \text{HCONH}_2 \leftrightharpoons \text{NH}_3 + \text{CO} \]With \( K_c = 4.84 \) at 400 K, the concentrations of the products (NH3 and CO) and reactants (HCONH2) can be determined.
Understanding the equilibrium constant is crucial because it indicates the extent of the reaction at equilibrium. A higher \( K_c \) means more products are formed.

Dissociation Reaction

Dissociation reactions involve the breakdown of a compound into two or more simpler substances. In the context of formamide, it dissociates into ammonia (\( \text{NH}_3 \)) and carbon monoxide (\( \text{CO} \)):
\[ \text{HCONH}_2(g) \rightarrow \text{NH}_3(g) + \text{CO}(g) \]The key to solving problems involving dissociation reactions is to set up an equilibrium table:

• Start with the initial concentrations.
• Determine changes in concentrations using \( x \) to denote the amount reacted.
• At equilibrium, calculate the concentrations using these changes.

Each reactant and product's concentration depends on the amount of \( x \) that moves from reactants to products in reaching equilibrium.
This concept is pivotal for understanding dynamic systems where reactants and products continuously interconvert until equilibrium is achieved.

Formamide Decomposition

Formamide (\( \text{HCONH}_2 \)) decomposes into ammonia and carbon monoxide at high temperatures. This process is significant in various industrial applications, such as pharmaceuticals and dyes.
During the decomposition in a closed system, the forward reaction where formamide breaks down occurs simultaneously with the reverse reaction where products revert to formamide.
The decomposition of formamide is represented by the equation:\[ \text{HCONH}_2(g) \rightleftharpoons \text{NH}_3(g) + \text{CO}(g) \]Important points to remember:

• As temperature increases, decomposition tends to favor the formation of products.
• Using the given \( K_c \), the amount of each gas present at equilibrium can be determined.

The understanding of this concept is vital for predicting and controlling the conditions under which formamide decomposes. It links the physical properties of gases with chemical equilibria to give insights into practical chemical processes.