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The equilibrium constant value for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) is \(1.48 \times 10^{-5}\), the value for the reaction \(1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+3 / 2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})\) is \(\mathrm{n} \times 10^{-3}\) where \(\mathrm{n}\) is (a) \(1.85\) (b) \(3.85\) (c) \(4.85\) (d) 10

Short Answer

Expert verified
\( n \approx 1.22 \), none of the options match exactly.

Step by step solution

01

Understand the Reaction Equilibrium

The given reaction is \( \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \). Its equilibrium constant is \( K_c = 1.48 \times 10^{-5} \). We need to find the equilibrium constant for the modified reaction \( \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g}) \).
02

Adjust the Reaction Coefficients

The new reaction is derived by dividing every coefficient in the original reaction by 2. The new reaction becomes \( \frac{1}{2} \mathrm{~N}_{2} + \frac{3}{2} \mathrm{H}_{2} \rightleftharpoons \mathrm{NH}_{3} \).
03

Use the Relationship between Equilibrium Constants

When a reaction equation is divided by a number, the equilibrium constant for the new equation is the square root of the original equation's equilibrium constant. Thus, the new \( K_c' \) is \( \sqrt{1.48 \times 10^{-5}} \).
04

Calculate the New Equilibrium Constant

Compute \( \sqrt{1.48 \times 10^{-5}} \). Using a calculator, \( \sqrt{1.48} \approx 1.216 \), and \( \sqrt{10^{-5}} = 10^{-2.5} = 0.001 \). Thus, \( K_c' = 1.216 \times 10^{-2.5} \) or \( 1.216 \times 0.001 \).
05

Determine "n" Value

After calculating, \( K_c' = 1.216 \times 0.001 = 0.001216 \). Hence, \( K_c' = 1.216 \times 10^{-3} \). Therefore, \( n = 1.216 \). None of the given options exactly match this value; thus, there might be an error in choices matching the calculation, or interpretations might differ slightly.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reaction Equilibrium
In chemical reactions, equilibrium refers to the state where the reactants and products are present in concentrations that do not change over time. This happens when the rate of the forward reaction equals the rate of the reverse reaction. This balance is what we call the **reaction equilibrium**.
Understanding how equilibrium works helps us predict the concentrations of products and reactants during the reaction process. In equilibrium, concentration measurements remain consistent, which is crucial in industries like pharmaceuticals where specific concentrations are needed.
One vital aspect is the dynamic nature of equilibrium, meaning molecules continue to react even though their overall concentrations stay the same. Equilibrium doesn't mean the reactants and products are in equal amounts; it’s about their rates balancing out. This dynamic balance can shift if conditions like pressure, temperature, or concentration change, a principle called Le Chatelier’s Principle.
Chemical Reactions
Chemical reactions involve the transformation of one or more substances into a new substance. Understanding the **chemical reactions** is crucial for calculating equilibrium constants. In chemical equations, reactants are written on the left and products on the right, separated by an arrow which indicates the direction of the reaction.
Like in our exercise with ammonia production, it's vital to consider the following:
  • Reactants: The starting materials \(\ce{N2} \text{and} \ce{H2}\).
  • Products: The outcome of the reaction \(\ce{NH3}\).
  • Coefficients: Numbers before molecules, indicating moles involved in the reaction (1 mole of \(\ce{N2}\) reacts with 3 moles of \(\ce{H2}\) to produce 2 moles of \(\ce{NH3}\).
These coefficients play a significant role in determining how much product is formed from given reactants under equilibrium conditions. Chemical reactions may reach equilibrium without all reactants converting to products, which underscores the role of equilibrium constants in predicting reaction behaviors.
Equilibrium Calculations
**Equilibrium calculations** allow us to predict how a chemical reaction behaves when it reaches equilibrium. The equilibrium constant ( K_c ) is a key part of these calculations and is a ratio of the concentrations of products to reactants, each raised to the power of their coefficients from the balanced equation.
In our exercise, to find the new equilibrium constant for a different reaction setup, we adjusted the coefficients of the original reaction. Changing a reaction’s coefficients alters the calculation of the equilibrium constant. For example:
  • If a reaction is halved, then the equilibrium constant of the new reaction becomes the square root of the original equilibrium constant.
  • If a reaction is reversed, the equilibrium constant becomes the reciprocal.
This stems from a fundamental property of equilibrium constants: they depend on the stoichiometry of the reaction, rather than time or initial concentrations. Calculating equilibria helps chemists determine conditions for processes to yield the maximum desired product, making it invaluable for manufacturing and research fields.

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Most popular questions from this chapter

A vessel at equilibrium, contains \(\mathrm{SO}_{3}, \mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\), Now some helium gas is added, so that total pressure increases while temperature and volume remain constant. According to Le Chatelier's Principle, the dissociation of \(\mathrm{SO}_{3}\) (a) decreases (b) remains unaltered (c) increases (d) change unpredictably

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A saturated solution of non-radioactive sugar was taken and a little radioactive sugar was added to it. A small amount of it gets dissolved in solution and an equal amount of sugar was precipitated. This proves (a) the equilibrium has been established in the solution (b) radioactive sugar can displace non-radioactive sugar from its solution. (c) Equilibrium is dynamic in nature (d) none of the above

Which of the following reaction will be favoured at low pressure? (a) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) (b) \(\mathrm{H}_{2}+\mathrm{I}_{2}=2 \mathrm{HI}\) (c) \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\) (d) \(\mathrm{N}_{2}+\mathrm{O}_{2}=2 \mathrm{NO}\)

In a reaction \(\mathrm{A}_{2}(\mathrm{~g})+4 \mathrm{~B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}_{4}(\mathrm{~g}) ; \Delta \mathrm{H}<0\). The formation of \(\mathrm{AB}_{4}(\mathrm{~g})\) will be favoured by (a) low temperature and high pressure (b) high temperature and high pressure (c) low temperature and low pressure (d) high temperature and low pressure

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