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Vilka [71]
3 years ago
7

Balance the chemical equation below: CO2 + H2O + 2678 kJ-> C6H1206 + O2 a. How many moles of H20 are involved in this reactio

n? b. Is this reaction Exothermic or Endothermic? c. How much energy will be involved in this reaction when 1 mole of CO2 is used up?
Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0

Answer:

a) 6 mol H2O

b) this reaction is endothermic

c) when 1 mol of CO2 is used, in the reaction they occur 0.5025 KJ

Explanation:

balanced eq:

  • 6CO2 + 6H2O + 2678 KJ ↔ C6H12O6 + 6O2

                                         6 - C - 6

                                         18 - O - 18

                                          12 - H - 12

a) mol H2O = 6 mol.......from balanced equation.

b) ΔE = 2678 KJ....... this reaction absorbs heat ( ΔE is positive )

c)    1 gramo C6H12O6 ≅ 4 cal

  • Mw C6H12O6 = 180.156 g/mol

⇒ 1mol CO2 * ( mol C6H12O6 / 6mol CO2 ) =0.166 mol C6H12O6

⇒ 0.166mol C6H12O6 * ( 180.156 g C6H12O6 / mol ) = 30.026g C6H12O6

⇒30.026 gC6H12O6 * ( 4 cal / gC6H12O6 ) * ( Kcal / 1000 cal ) * (4184 J / Kcal ) * ( KJ / 1000 J ) = 0.5025 KJ C6H12O6.

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The following sequence of reactions occurs in the commercial production of aqueous nitric acid: 4NH3(g)+5O2(g)⟶4NO(g)+6H2O(l)ΔH=
castortr0y [4]

Answer:

The total energy change for the production of one mole of aqueous nitric acid is −494 kJ

Explanation:

<u>Step 1</u>: Data given

4NH3(g)+5O2(g)⟶4NO(g)+6H2O(l)ΔH=−907kJ 2NO(g)+O2(g)⟶2NO2(g)ΔH=−113kJ 3NO2+H2O(l)⟶2HNO3(aq)+NO(g)ΔH=−139kJ

<u>Step 2:</u> Multiply equations

Multiply the first equation by 3:

12 NH3(g) + 15 O2(g) → 12 NO(g) + 18 H2O(l) ΔH = −2721 kJ

Multiply the second equation by 6:

12 NO(g) + 6 O2(g) → 12 NO2(g) ΔH = −678 kJ

Multiply the third equation by 4:

12 NO2(g) + 4 H2O(l) → 8 HNO3(aq) + 4 NO(g) ΔH = −556 kJ

<u>Step 3:</u> Get the equations together

12 NH3(g) + 15 O2(g) + 12 NO(g) + 6 O2(g) + 12 NO2(g) + 4 H2O(l) →

12 NO(g) + 18 H2O(l) + 12 NO2(g) + 8 HNO3(aq) + 4 NO(g)

ΔH = −2721 kJ − 678 kJ − 556 kJ

We can simplify as followed:

12 NH3(g) + 21 O2(g) → 14 H2O(l) + 8 HNO3(aq) + 4 NO(g) ΔH = −3955 kJ

<u> Step 4:</u> Determine the total energy change for the production of one mole of aqueous nitric acid by this process:

−3955 kJ/8 moles HNO3= −494 kJ

The total energy change for the production of one mole of aqueous nitric acid is −494 kJ

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3 years ago
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