The standard reaction free energy [tex]$\Delta G^0=70 . kJ$[/tex] for this reaction:
[tex]$MgCl_2(s)+H_2 O(l) \rightarrow MgO(s)+2 HCl(g)$[/tex]
Use this information to complete the table below. Round each of your answers to the nearest kJ.
| reaction | [tex]$\Delta G^0$[/tex] |
|---|---|
| [tex]$MgO (s)+2 HCl (g) \rightarrow MgCl _2(s)+ H _2 O (l)$[/tex] | kJ |
| [tex]$\frac{1}{2} MgO ( s )+ HCl ( g ) \rightarrow \frac{1}{2} MgCl _2(s)+\frac{1}{2} H _2 O ( l )$[/tex] | kJ |
| [tex]$\frac{1}{4} MgCl _2(s)+\frac{1}{4} H _2 O ( l ) \rightarrow \frac{1}{4} MgO ( s )+\frac{1}{2} HCl ( g )$[/tex] | 18 kJ |
Answer (2)
When the reaction is reversed, the sign of Δ G 0 changes: Δ G 0 = − 70 kJ .
For half of the reverse reaction, Δ G 0 is halved: Δ G 0 = 2 1 ( − 70 kJ ) = − 35 kJ .
For one-fourth of the original reaction, Δ G 0 is one-fourth: Δ G 0 = 4 1 ( 70 kJ ) = 17.5 kJ ≈ 18 kJ .
The final answers are: − 70 , − 35 , and 18 kJ.
Explanation
Problem Analysis We are given the standard Gibbs free energy change, Δ G 0 = 70 kJ , for the reaction: M g C l 2 ( s ) + H 2 O ( l ) → M g O ( s ) + 2 H Cl ( g ) . We need to find the Δ G 0 for three related reactions.
Reverse Reaction The first reaction is the reverse of the given reaction: M g O ( s ) + 2 H Cl ( g ) → M g C l 2 ( s ) + H 2 O ( l ) . When a reaction is reversed, the sign of Δ G 0 changes. Therefore, for this reaction, Δ G 0 = − 70 kJ .
Half Reverse Reaction The second reaction is: 2 1 M g O ( s ) + H Cl ( g ) → 2 1 M g C l 2 ( s ) + 2 1 H 2 O ( l ) . This reaction is half of the reverse of the given reaction. Therefore, the Δ G 0 for this reaction is half of the Δ G 0 for the reverse reaction: Δ G 0 = 2 1 ( − 70 kJ ) = − 35 kJ .
Quarter Reaction The third reaction is: 4 1 M g C l 2 ( s ) + 4 1 H 2 O ( l ) → 4 1 M g O ( s ) + 2 1 H Cl ( g ) . The Δ G 0 for this reaction is given as 18 kJ. We can verify this by taking one-fourth of the original reaction's Δ G 0 : 4 1 ( 70 kJ ) = 17.5 kJ . Rounding to the nearest kJ, we get 18 kJ.
Final Answer Now, we can fill in the table with the calculated values.
Examples
Gibbs free energy is a crucial concept in chemistry, helping us predict the spontaneity of reactions. For instance, consider designing a new industrial process to produce a chemical compound. By calculating the Gibbs free energy change for each step of the process, chemists can determine whether the reaction will proceed spontaneously under given conditions. If the Gibbs free energy change is negative, the reaction is likely to occur without external energy input, making the process more efficient and cost-effective. This principle is also vital in developing new materials, optimizing energy storage solutions, and understanding biological processes.
The calculated ΔG° values are -70 kJ for the reverse reaction, -35 kJ for the half reverse reaction, and 18 kJ for the quarter reaction. These values help to understand the Gibbs free energy changes associated with these reactions. It demonstrates how reversing or scaling reactions affects their thermodynamic properties.
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