Select the correct answer.
What would cause the equilibrium to shift left in this reaction?
[tex]$CO+3 H_2 \rightleftharpoons CH_4+H_2 O$[/tex]
A. adding heat to the product mixture
B. a faster rate of forward reaction
C. increased collisions between CO and [tex]$H _2$[/tex]
D. the escape of water from the mixture
E. placing the mixture in a cold water bath
Answer (1)
The reaction is exothermic ( Δ H < 0 ).
Adding heat shifts the equilibrium to the left.
Other options either do not shift the equilibrium or shift it to the right.
Therefore, the correct answer is adding heat to the product mixture.
A
Explanation
Understanding the Reaction and Equilibrium Let's analyze the given chemical reaction and the factors that can shift its equilibrium. The reaction is: CO ( g ) + 3 H 2 ( g ) ⇌ C H 4 ( g ) + H 2 O ( g ) The question asks us to identify which of the provided options would cause the equilibrium to shift to the left, meaning favoring the reactants (CO and H 2 ). We'll use Le Chatelier's principle to evaluate each option.
Determining Enthalpy Change First, we need to determine whether the reaction is exothermic or endothermic. We can calculate the enthalpy change ( Δ H ) using standard enthalpies of formation. The values are:
CO ( g ) : -110.5 kJ/mol
H 2 ( g ) : 0 kJ/mol
C H 4 ( g ) : -74.6 kJ/mol
H 2 O ( g ) : -241.8 kJ/mol
The enthalpy change of the reaction is calculated as: Δ H = [ Δ H f ( C H 4 ) + Δ H f ( H 2 O )] − [ Δ H f ( CO ) + 3 × Δ H f ( H 2 )] Δ H = [( − 74.6 ) + ( − 241.8 )] − [( − 110.5 ) + 3 ( 0 )] Δ H = − 316.4 + 110.5 = − 205.9 kJ/mol Since Δ H < 0 , the forward reaction is exothermic, meaning it releases heat.
Analyzing Each Option Now, let's analyze each option:
A. Adding heat to the product mixture: Since the forward reaction is exothermic, adding heat will shift the equilibrium to the left, favoring the reactants. This is because the system will try to counteract the increase in heat by consuming some of the added heat, which is achieved by shifting the equilibrium towards the reactants.
B. A faster rate of forward reaction: A faster rate of forward reaction does not necessarily shift the equilibrium. It only means the equilibrium is reached faster. The equilibrium position remains the same.
C. Increased collisions between CO and H 2 : Increased collisions between reactants would favor the forward reaction, shifting the equilibrium to the right, not the left.
D. The escape of water from the mixture: The escape of water from the mixture would decrease the concentration of products, shifting the equilibrium to the right to compensate for the loss of product.
E. Placing the mixture in a cold water bath: Placing the mixture in a cold water bath is equivalent to removing heat. Since the forward reaction is exothermic, removing heat will shift the equilibrium to the right, favoring the products.
Conclusion Based on our analysis, adding heat to the product mixture (Option A) will cause the equilibrium to shift to the left. This is because the reaction is exothermic, and adding heat favors the reverse reaction to consume the excess heat.
Examples
Consider an industrial process where methane ( C H 4 ) is produced from carbon monoxide (CO) and hydrogen ( H 2 ). Understanding how to shift the equilibrium of this reaction is crucial for optimizing methane production. If the goal is to maximize methane production, the reaction conditions should be adjusted to favor the forward reaction (e.g., by removing heat). Conversely, if the goal is to reduce CO levels, conditions should be adjusted to favor the reverse reaction (e.g., by adding heat). This principle is widely applied in chemical industries to control reaction outcomes and improve efficiency.
