2CO(g) + O\(_2\)(g) → 2CO\(_2\)(g)
Given that ΔH[CO]is -110.4 kJmol-1 and ΔH[CO\(_2\)] is -393.0 kJmol-1, the energy change for the reaction above is

a

-503.7 kJ

b

-282.6 kJ

c

+282.6 kJ

d

+503.7 kJ

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Explanation

Correct Option
b

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Discussions (21)

marlianlawa
2 years ago

Answer should be -565.2. Those Heat Change values are for 1 mole not 2.

OgeJesus
4 years ago

please how is the 7 gotten

Chellie31
6 years ago

Where did the 7 in the explanation come from?

joyiember
10 years ago

The selected answer is wrong:

Your explanation has a different answer to the selected one

Myschool Blessing
10 years ago

Thanks for your contributions. Corrections have been made.

FamDamSam
12 years ago

Regarding the correction, the correct option is option B

gabby12
11 years ago

The correct answer to the question is option B.check your correction well.

Chukwu3m3ri3
2 years ago

πŸ–•

tobilad
2 years ago

The balanced equation is: 2CO(g) + O2(g) β†’ 2CO2(g)
(2 Γ— -393.0 kJ/mol) - (2 Γ— -110.4 kJ/mol)
Ξ”H = -503.6 kJ/mol

BrownStanley2007
2 years ago

Apologies for the confusion. Let's clarify the correct calculation:

We have already determined that the enthalpy change for the reaction (βˆ†H_rxn) is -565.2 kJ/mol.

To find the total energy change for the reaction, we need to consider the coefficients in the balanced chemical equation. The coefficients indicate the stoichiometry of the reaction.

In the given reaction:
2CO(g) + O2(g) β†’ 2CO2(g)

The coefficients are:
2 mol of CO reacts with 1 mol of O2 to produce 2 mol of CO2.

So, for every 2 mol of CO reacting, the enthalpy change is -565.2 kJ/mol.

But if we consider 1 mol of CO reacting, the enthalpy change would be half of -565.2 kJ/mol since the coefficients are halved.

Hence, the correct answer is:

-565.2 kJ/mol / 2 = -282.6 kJ/mol

Therefore, the correct answer is B. -282.6 kJ.

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