The equilibrium constant, K., for the following reaction is 1.96×10° at 548 K. NH3(g) + HCI(g) NHẠCI(s) This reaction is favored at equilibrium. REACTANT Enter PRODUCT or REACTANT. The concentrations of NH3 and HCl will be LOW at equilibrium. Enter HIGH or LOW.
Basics in Organic Reactions Mechanisms
In organic chemistry, the mechanism of an organic reaction is defined as a complete step-by-step explanation of how a reaction of organic compounds happens. A completely detailed mechanism would relate the first structure of the reactants with the last structure of the products and would represent changes in structure and energy all through the reaction step.
Heterolytic Bond Breaking
Heterolytic bond breaking is also known as heterolysis or heterolytic fission or ionic fission. It is defined as breaking of a covalent bond between two different atoms in which one atom gains both of the shared pair of electrons. The atom that gains both electrons is more electronegative than the other atom in covalent bond. The energy needed for heterolytic fission is called as heterolytic bond dissociation energy.
Polar Aprotic Solvent
Solvents that are chemically polar in nature and are not capable of hydrogen bonding (implying that a hydrogen atom directly linked with an electronegative atom is not found) are referred to as polar aprotic solvents. Some commonly used polar aprotic solvents are acetone, DMF, acetonitrile, DMSO, etc.
Oxygen Nucleophiles
Oxygen being an electron rich species with a lone pair electron, can act as a good nucleophile. Typically, oxygen nucleophiles can be found in these compounds- water, hydroxides and alcohols.
Carbon Nucleophiles
We are aware that carbon belongs to group IV and hence does not possess any lone pair of electrons. Implying that neutral carbon is not a nucleophile then how is carbon going to be nucleophilic? The answer to this is that when a carbon atom is attached to a metal (can be seen in the case of organometallic compounds), the metal atom develops a partial positive charge and carbon develops a partial negative charge, hence making carbon nucleophilic.
dont know why im wrong
![### Equilibrium Constants and Chemical Reactions: An Example
For the reaction:
\[ \text{NH}_3(\text{g}) + \text{HCl}(\text{g}) \leftrightarrow \text{NH}_4\text{Cl}(\text{s}) \]
the equilibrium constant, \( K_c \), is valued at \( 1.96 \times 10^5 \) at a temperature of 548 K.
**Understanding the Reaction:**
1. **Reaction Direction:**
- This reaction is **product** favored at equilibrium. (Enter PRODUCT or REACTANT)
2. **Concentration of Reactants:**
- The concentrations of \( \text{NH}_3 \) and \( \text{HCl} \) will be **low** at equilibrium. (Enter HIGH or LOW)
### Explanation:
**Equilibrium Constant (\( K_c \)):**
- A large equilibrium constant (\( K_c \)) signifies that the products are highly favored when the reaction is at equilibrium.
- In this example, \( K_c = 1.96 \times 10^5 \), which is quite large, indicating that at equilibrium, most of the reactants \( \text{NH}_3 \) and \( \text{HCl} \) have been converted to the product \( \text{NH}_4\text{Cl} \).
**Concentration Analysis:**
- Given that the reaction is product-favored, the equilibrium concentrations of the reactants \( \text{NH}_3 \) and \( \text{HCl} \) will be low, as most of them will have reacted to form \( \text{NH}_4\text{Cl} \).
This behavior aligns with Le Chatelier's principle which states that the system will adjust itself to counter any changes imposed on it, thus driving the reaction towards product formation given the large equilibrium constant.
### Conclusion:
Understanding the equilibrium constant helps in predicting the position of equilibrium and the relative concentrations of reactants and products. For students learning chemistry, grasping such fundamentals is key to mastering reaction dynamics and chemical equilibrium concepts.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbc7db68c-7828-4a06-bc34-18cdd71bf49b%2F4070b081-d2bf-4d1d-96d3-9e5b9b40c678%2Fcvotj5o_processed.png&w=3840&q=75)
Step by step
Solved in 2 steps with 2 images
