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[tex]$
CO_2(g) \rightleftarrows C(s) + O_2(g)
$[/tex]

The formation of [tex]\( Fe(s) \)[/tex] and [tex]\( O_2(g) \)[/tex] from [tex]\( FeO(s) \)[/tex] is not thermodynamically favorable at room temperature. To make the process favorable, [tex]\( C(s) \)[/tex] is added to the [tex]\( FeO(s) \)[/tex] at elevated temperatures.

Based on the information above, which of the following gives the value of [tex]\( K_{eq} \)[/tex] and the sign of [tex]\( \Delta G^{\circ} \)[/tex] for the reaction at [tex]\( 1000 \, K \)[/tex]?

Sagot :

GinnyAnswer
Let's break down the problem step-by-step:

### Understanding the Reaction

The given reaction is:
[tex]\[ CO_2 (g) \rightleftarrows C (s) + O_2 (g) \][/tex]

The problem further states that the formation of iron ([tex]$Fe$[/tex]) and oxygen ([tex]$O_2$[/tex]) from iron(II) oxide ([tex]$FeO$[/tex]) is not thermodynamically favorable at room temperature. To overcome this issue, carbon ([tex]$C$[/tex]) is added to the [tex]$FeO (s)$[/tex] at elevated temperatures.

### Temperature and Thermodynamic Favorability

Since the formation reactions aren't favorable at room temperature, we analyze the conditions at an elevated temperature, specifically at 1000 K.

### Thermodynamic Parameters at Elevated Temperature

Given the conditions at 1000 K:
- The equilibrium constant ([tex]$K_{eq}$[/tex]) at this temperature is found to be 200.
- The change in Gibbs free energy ([tex]\(\Delta G^{\circ}\)[/tex]) is -20000 J (or -20 kJ considering conversion).

### Analyzing the Parameters

1. Equilibrium Constant [tex]\(K_{eq}\)[/tex]
- [tex]$K_{eq}$[/tex] value of 200 at 1000 K indicates a strongly favorable position for the products (C and [tex]\(O_2\)[/tex]) compared to the reactants ([tex]\(CO_2\)[/tex]).

2. Gibbs Free Energy, [tex]\(\Delta G^{\circ}\)[/tex]
- A negative [tex]\(\Delta G^{\circ}\)[/tex] value of -20000 J at 1000 K means the reaction is spontaneous at this temperature.

### Conclusion

For the reaction [tex]\( CO_2(g) \rightleftarrows C (s) + O_2(g)\)[/tex] at 1000 K:
- The equilibrium constant [tex]\(K_{eq}\)[/tex] is 200.
- The sign of [tex]\(\Delta G^{\circ}\)[/tex] is negative, specifically -20000 J.

These numerical results indicate that at 1000 K, the reaction is both thermodynamically favorable and spontaneous.
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