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What is the unit of \(K_{p}\) for the reaction? $$ \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) $$ (a) atm (b) \(\mathrm{atm}^{-2}\) (c) \(\mathrm{atm}^{2}\) (d) \(\mathrm{atm}^{-1}\)

Short Answer

Expert verified
The unit of \(K_p\) for the given reaction is \(\mathrm{atm}^{-1}\).

Step by step solution

01

Understand The Balanced Chemical Equation

The balanced chemical equation given is: \[\mathrm{CS}_2(g) + 4\mathrm{H}_2(g) \rightleftharpoons \mathrm{CH}_4(g) + 2\mathrm{H}_2 \mathrm{S}(g)\]. This shows one mole of carbon disulfide and four moles of hydrogen gas reacting to form one mole of methane and two moles of hydrogen sulfide.
02

Write the Expression for Kp

The equilibrium constant for pressure, \(K_p\), is expressed in terms of the partial pressures of the gases. For the given reaction, the expression is: \[K_p = \frac{P(\mathrm{CH}_4) \cdot P(\mathrm{H}_2S)^2}{P(\mathrm{CS}_2) \cdot P(\mathrm{H}_2)^4}\].
03

Determine the Units for Kp

The unit for partial pressures is atmospheres (atm). Therefore, to determine the units for \(K_p\) we simply insert the units into the expression:\[K_p = \frac{(\mathrm{atm}) \cdot (\mathrm{atm})^2}{(\mathrm{atm}) \cdot (\mathrm{atm})^4} = \mathrm{atm}^{-1}\]. When the units are combined, they simplify to \(\mathrm{atm}^{-1}\), which is the unit for \(K_p\) for this reaction.

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Key Concepts

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

Chemical Equilibrium
When a chemical reaction takes place, reactants are converted into products, but under certain conditions, the reverse reaction can occur simultaneously, leading to a state called chemical equilibrium. This is a dynamic state where the rate of the forward reaction equals the rate of the backward reaction, meaning that the concentration of reactants and products remains constant over time. In the given exercise, carbon disulfide and hydrogen gas are in equilibrium with methane and hydrogen sulfide gas. Such equilibria are well-described by the equilibrium constant, which in the context of gaseous reactions and partial pressures, is denoted as \(K_p\).
Understanding the concept of equilibrium is crucial because it tells us that at a given temperature, regardless of how much we start with, the ratio of the concentration of products to reactants, raised to the power of their coefficients in the balanced equation, reaches a constant value. This concept links to how we predict the extent of a reaction, and whether a system favors products or reactants under specified conditions.
Partial Pressure
The term partial pressure refers to the pressure exerted by an individual gas in a mixture of gases. It is directly proportional to the mole fraction of that gas in the mixture and is a way to express the concentration of gas in terms of pressure. For the reaction \(\mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+2\mathrm{H}_{2} \mathrm{S}(g)\), each gas exerts its own partial pressure.
In the context of chemical equilibrium for gases, partial pressures are used in calculating the equilibrium constant, \(K_p\). This is because gases mix uniformly and their behavior often follows the ideal gas law, making pressure a convenient measure of concentration. The knowledge of partial pressures is essential for understanding how changes in pressure can shift the equilibrium position, a concept known as Le Chatelier's principle.
Mole Concept
The mole concept is a fundamental aspect of chemistry that provides a bridge between the microscopic world of atoms and molecules and the macroscopic world of grams and liters. A mole represents 6.022 x 10^23 particles – whether atoms, molecules, or ions – and is the SI unit for amount of substance.
When dealing with reactions involving gases, the mole concept allows us to relate volume, pressure, and temperature via the ideal gas law. In our exercise, one mole of \(\mathrm{CS}_{2}\) reacts with four moles of \(\mathrm{H}_{2}\), resulting in one mole of \(\mathrm{CH}_{4}\) and two moles of \(\mathrm{H}_{2}S\). This molar ratio is vital for writing the \(K_p\) expression, where the coefficients in the balanced equation determine the exponents on the partial pressures. The mole concept underpins stoichiometry, thereby providing a quantitative description of the relationships between reactants and products in a chemical reaction.

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Most popular questions from this chapter

Some inert gas is added at constant volume to the following reaction at equilibrium $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ Predict the effect of adding the inert gas: (a) the equilibrium shifts in the forward direction (b) the equilibrium shifts in the backward direction (c) the equilibrium remains unaffected (d) the value of \(K_{p}\) is increased

A system at equilibrium is described by the equation of fixed temperature \(T\). $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ What effect will an increases in the total pressure caused by a decrease in volume have on the equilibrium?

A reversible reaction is one which : (a) Proceeds in one direction (b) Proceeds in both directions (c) Proceeds spontaneously (d) All the statements are wrong

The vapour pressure of a liquid in a closed container depends on : (1) temperature of liquid (2) quantity of liquid (3) surface area of the liquid (a) 1 only (b) 2 only (c) 1 and 3 only (d) 1,2 and 3

For which of the following reaction is product formation favoured by low pressure and low temperature? (a) \(\mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g) ; \quad \Delta H^{\circ}=172.5 \mathrm{~kJ}\) (b) \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}\) \(\Delta H^{\circ}=-21.7 \mathrm{~kJ}\) (c) \(2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\) \(\Delta H^{\circ}=-285 \mathrm{~kJ}\) (d) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) \(\Delta H^{\circ}=-541 \mathrm{~kJ}\)

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