Consider the following reaction occurring at 298 k: baco3(s)⇌bao(s)+co2(g)

Think about the after effect
happening at 298 K: BaCO3(s)⇌BaO(s)+CO2(g) think about the after effect
happening at 298 K: BaCO3(s)⇌BaO(s)+CO2(g)
A) Show that effect isn’t natural under standard
problems by determining ΔG∘rxn. G B) If BaCO3 is put in an evacuated flask, exactly what limited stress
of CO2 should be current as soon as the effect hits balance? C) exactly what heat is needed to create a carbon dioxide partial
stress of 1.0 atm?

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January 1, 2022

Consider the following reaction occurring at 298 K: BaCO3(s)⇌BaO(s)+CO2(g)
A) Show that the reaction is not spontaneous under standard conditions by calculating ΔG∘rxn.

B) If BaCO3 is placed in an evacuated flask, what partial pressure of CO2 will be present when the reaction reaches equilibrium?

C) What temperature is required to produce a carbon dioxide partial pressure of 1.0 atm?

Answer

Question

Consider the following reaction occurring at 298 K: BaCO3(s)⇌BaO(s)+CO2(g) Part A Show that the reaction is...

Consider the following reaction occurring at 298 K:
BaCO3(s)⇌BaO(s)+CO2(g)

Part A

Show that the reaction is not spontaneous under standard conditions by calculating ΔG∘rxn.

ΔG∘rxn =?

Part B

If BaCO3 is placed in an evacuated flask, what partial pressure of CO2 will be present when the reaction reaches equilibrium?

P=?

Part C

What temperature is required to produce a carbon dioxide partial pressure of 1.0 atm?

T=?

Answers

PartA: The standard free energy change is positive at 298K and 1 bar, indicating that the reaction is not spontaneous under standard conditions and is therefore reactant-favored at equilibrium ΔGrem =[(-394.4)-(一5 20.3]-[1(-1134.4)] rexn -914.7 +1134.4 219.7kJ Since-Gorn 1s positive, the rreaction is Non spontaneous. Part B: BaCO, (s) Bao (s) CO. (g) K Pco RTlnHK RT 219.7k 88.67 K- -3.08x1089 atim Pco Part C: AHe -393.5)+172.1)]-[(-1213.0) - +271.5kJ 1213.)72.1-1(112.1) = +173 .8JK Since, AS is positive,so the reaction will become spontaeous at high T Therefore, +271.5kJ - 1562K AS

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