Question

When one mole of carbon disulfide gas reacts with hydrogen gas, methane and hydrogen sulfide gases are formed. When equilibrium is reached at \(900^{\circ} \mathrm{C}\), analysis shows that \(P_{\mathrm{CH}_{4}}=0.0833 \mathrm{~atm}, P_{\mathrm{H}_{2} \mathrm{~s}}=0.163 \mathrm{~atm}, P_{\mathrm{CS}_{2}}=\) \(1.27 \mathrm{~atm},\) and \(P_{\mathrm{H}_{2}}=0.873 \mathrm{~atm}\) (a) Write a balanced equation (smallest whole-number coefficients) for the reaction. (b) Find \(K\) at \(900^{\circ} \mathrm{C}\).

Short answer

Question: Determine the equilibrium constant, Kp, for the reaction between carbon disulfide and hydrogen gas to form methane and hydrogen sulfide at 900°C, given the following partial pressures at equilibrium: \(P_{CH_4} = 0.0833\ atm\), \(P_{H_2S} = 0.163\ atm\), \(P_{CS_2} = 1.27\ atm\), and \(P_{H_2} = 0.873\ atm\). Answer: The equilibrium constant, Kp, for the reaction under the given conditions is approximately 0.053.

Step-by-step solution

Write the balanced chemical equation

To write the balanced equation, we need to make sure that the number of atoms on each side of the equation is equal. Carbon disulfide (CS2) reacts with Hydrogen (H2) to form Methane (CH4) and Hydrogen sulfide (H2S). So the balanced equation will be: \(CS_2 + 4H_2 \rightarrow CH_4 + 2H_2S\)

Write the expression for the equilibrium constant

The equilibrium constant K can be defined in terms of partial pressures for gas phase reactions. The expression for the equilibrium constant, Kp, in terms of partial pressures for the given balanced equation is as follows: \(K_p = \frac{P_{CH_4} \cdot P_{H_2S}^2}{P_{CS_2} \cdot P_{H_2}^4}\)

Substitute the given values into the equation and calculate Kp

We are given the partial pressures at equilibrium: \(P_{CH_4} = 0.0833\ atm\), \(P_{H_2S} = 0.163\ atm\), \(P_{CS_2} = 1.27\ atm\), and \(P_{H_2} = 0.873\ atm\). Substitute these values into the Kp expression: \(K_p = \frac{0.0833 \cdot (0.163)^2}{1.27 \cdot (0.873)^4}\) After performing the calculations, we find the equilibrium constant: \(K_p ≈ 0.053\) So, the equilibrium constant for this reaction under the given conditions is \(K_p ≈ 0.053\).

Key concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant is a crucial concept. It is represented by the symbol 'K' and provides insight into the relationship between products and reactants when the reaction has reached a state where the concentrations no longer change with time, known as chemical equilibrium. For gas phase reactions, we often use the symbol 'K_p' to denote the equilibrium constant in terms of partial pressures.

The general form of the equilibrium expression for a reaction, such as A + bB \rightarrow cC + dD , with lowercase letters representing stoichiometric coefficients, is given by \[K_p = \frac{{P_C^c \cdot P_D^d}}{{P_A^a \cdot P_B^b}}\].

Each partial pressure (P) is raised to the power of its respective coefficient in the balanced equation. A higher value of 'K' suggests that, at equilibrium, the reaction favors the formation of products, while a lower 'K' indicates that the reactants are more favored. Understanding the equilibrium constant is essential for predicting the extent and direction of chemical reactions under varying conditions.

Partial Pressures

' is the total pressure of the gas mixture. By understanding and applying the concept of partial pressures, students can solve equilibrium problems for systems involving gases, as illustrated in the equilibrium constant calculation from the original exercise.

Balanced Chemical Equations

from our original exercise shows that one carbon atom and four hydrogen atoms react to form one methane molecule and two molecules of hydrogen sulfide. This balance is essential not only for the stoichiometry of the reaction - determining the relative amounts of reactants and products - but also for calculating the equilibrium constant, where the balanced equation determines the exponents in the equilibrium expression. Without a correctly balanced equation, the calculated equilibrium constant would not be accurate, which could lead to misunderstandings about the reaction's behavior.