Compare your equilibrium constant expressions from Problem 14.48(a) and Problem 14.49.

(a) Explain how the value of K_eq changes for a reaction when you double the reaction (when you multiply it through by 2).

(b) Suppose K_eq = 10.0 for a reaction. What will the value of K_eq be for the “doubled” reaction?

Data from Problem 14.49

Take the chemical reaction in Problem 14.48 and multiply it through by 2 (this is essentially like doubling a recipe). Write the new chemical reaction and the equilibrium constant expression for the doubled reaction.

Problem 14.48

Consider the gas-state reaction

Write the equilibrium constant expression for the reaction.

Write the reaction in reverse.

Write the equilibrium constant expression for the reverse reaction that you wrote for part (b).

Compare your answers to (a) and (c). What conclusion can you draw from the comparison?

Suppose K_eq for a reaction =10.0. What will the value of K_eq be for the reverse reaction?

2 A(g) + 3 B(g) C(g) + D(g)