Consider the intermediate chemical reactions.

2 equations. First: upper C a (s) plus upper C upper O subscript 2 (g) plus one half upper O subscript 2 (g) right arrow upper C a upper C upper O subscript 3 (s). Delta H 1 equals negative 812.8 kilojoules. Second: 2 upper C a (s) plus upper O subscript 2 (g) right arrow 2 upper C a upper O (s). Delta H 2 equals negative 1, 269 kilojoules.

The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C upper O subscript 3 (s).. When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation

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29-05-2020
Chemistry

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Consider the intermediate chemical reactions.

2 equations. First: upper C a (s) plus upper C upper O subscript 2 (g) plus one half upper O subscript 2 (g) right arrow upper C a upper C upper O subscript 3 (s). Delta H 1 equals negative 812.8 kilojoules. Second: 2 upper C a (s) plus upper O subscript 2 (g) right arrow 2 upper C a upper O (s). Delta H 2 equals negative 1, 269 kilojoules.

The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C upper O subscript 3 (s).. When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation

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dsdrajlin dsdrajlin · Answer 1 · 2021-12-10T00:11:10+01:00

When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

Let's consider the following intermediate chemical reactions.

Reaction 1: Ca(s) + CO₂(g) + ½O₂(g) → CaCO₃(s) ΔH₁ = -812.8 kJ

Reaction 2: 2 Ca(s) + O₂(g) → 2 CaO(s) ΔH₂ = -1269 kJ

We want to calculate the enthalpy of the following overall chemical equation.

CaO(s) + CO₂(g) → CaCO₃(s) ΔH = ?

We can apply Hess' law.

What is Hess' law?

Hess' law states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.

To apply Hess' law in this case we have to:

Conserve equation 1 as is.
Reverse equation 2 and multiply it by 1/2, thus changing the sign of its enthalpy and multiplying it by 1/2.
Add both equations and their enthalpies.

Ca(s) + CO₂(g) + ½ O₂(g) → CaCO₃(s) ΔH = -812.8 kJ

+

CaO(s) → Ca(s) + ½ O₂(g) ΔH = 634.5 kJ

--------------------------------------------------------------------------------------

CaO(s) + CO₂(g) → CaCO₃(s) ΔH = -178.3 kJ

When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

The complete question is as follows.

Consider the intermediate chemical reactions.

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

The final overall chemical equation is

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?

When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation

is halved and has its sign changed.
is halved.
has its sign changed.
is unchanged.

Learn more about Hess' law here: https://brainly.com/question/11628413

s139297 s139297 · Answer 2 · 2021-12-11T13:36:26+01:00

s139297 s139297

11-12-2021

Answer:

i believe its A

Explanation:

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