banner

The water gas shift reaction CO(g)+H_2 O⇌ CO_2(g)+H_2(g) is used industrially to produce hydrogen. The reaction enthalpy is ΔH^o=-41kj. To increase the equilibrium yield of hydrogen would you use high or low temperature?

The question aims to determine whether the low or high temperature is favorable to increase the hydrogen yield for an exothermic reaction when a change in enthalpy is negative. The chemical equilibrium is not static but is naturally strong. At equilibrium, the forward and backward reactions continue at the same rate. Therefore, the production of reactants and products remains unchanged over time.

It can be confirmed by taking the example of the formation of ammonia from nitrogen and hydrogen. Under test conditions, we can assume the equivalent composition of $NH_{3}$, $N_{2}$, and $H_{2}$. Only nitrogen in this mixture absorbs radioactive nitrogen isotopes. All-natural reactions follow a standard process because they occur in a way that leads to a dramatic decline or increase in our natural energy.

When the total energy or temperature change increases or decreases, these two reactions are called exothermic or endothermic reactions. A chemical reaction in which heat escapes the system to an area is an exothermic reaction. In the thermodynamic sense, the temperature change is bad when heat comes out of the system.

A chemical reaction in which a system from the environment absorbs heat is called an endothermic reaction. According to the thermodynamic concept, when the system absorbs heat, the temperature conversion is good. Most exothermic reactions are reversible, but the reversal process should be natural endothermic.

For example, esterification of acetic acid in alcohol solution, ammonia concentration, and esterification of hydrocarbon amylene $(C_{5}H_{10})$. If the forward reaction in the process is endothermic, then the reversal process occurs with exothermic. Examples of endothermic reactions are melting ice to water, water vaporization, solid $Co_{2}$ sublimation, and bread baking.

Expert Answer

Let’s consider the following reaction:

\[CO{(g)}+H_{2}O \rightleftharpoons CO_{2}(g)+H_{2}(g)\]

$ \Delta H $ of this reaction is negative; if the measurement system is disrupted, the system response is defined by Le Chatelier’s Principle: If the measurement system suffers from a disturbance (temperature, pressure, concentration), the system will clear its measurement area to withstand that disturbance.

In this process, reaction is exothermic $(\Delta H^{o} < 0)$. We can consider heat as one of the products. When we lower the temperature, the system will try to raise it, allowing for a further reaction to release heat and increase $H_{₂}$ yield simultaneously. With more products, the value of permanent equilibrium increases.

Numerical Result

Reaction is an exothermic reaction $(\Delta H^{o} < 0)$. We can consider heat as one of the products. When we lower the temperature, the system will try to raise it, allowing for a further reaction to release heat and increase $H_{₂}$ yield simultaneously. With more products, the value of permanent equilibrium increases.

Example

$2NH_{3}\rightarrow N_{2}+3H_{2}\Delta H$

The change in enthalpy is

$\Delta H=+92.22\: kJ\: mol^{-1}$

To increase the production of $N_{2}$, will the increase or decrease of temperature become favorable?

Solution

The following reaction is endothermic because its change in enthalpy is positive. An increase in temperature is favorable to increase the production of $N_{2}$.

5/5 - (17 votes)