Calculate the enthalpy of reaction using bond energies given A) + 366 KJ B) -48.

Question

Calculate the enthalpy of reaction using bond energies given A) + 366 KJ B) -48.0 KJ C) -484 KJ D) +68.0 kJ E) -366 KJ Which of the following statements is are true: I. Outer electrons efficiently shield one another from nuclear charge. II. Core electrons effectively shield outer electrons from nuclear charge. III Valence electrons are most difficult of all electrons to remove. IV Core electrons have the lowest ionization energies of all electrons. A) II only B) I and III only C) IV only D) I only E) II and IV only

Explanation / Answer

17. CORE ELECTRONS ACT AS SCREENS AND PROTECT OUTER ELECTRONS FROM NUCLEAR CHARGE HENCE OUTER ELECTRONS ARE EASILY REMOVED THIS IS CALLED SCREENING EFFECT OR SHEILDING EFFECT. answer : II only

16. ENTHALPY OF REACTION

= SUM OF ENERGIES OF BONDS BROKEN - SUM OF ENERGIES OF BONDSFORMED

= BOND ENTHALPIES OF REACTANTS - BOND ENTHALPIES OF PRODUCTS

TWO C-C BONDS BROKEN = 2 X 347 kJ/mol

6 C-H BONDS ARE BROKEN = 6 X 414 kJ/mol

ONE C=O BOND IS BROKEN = 1X 745 kJ/mol

ONE H-H BOND IS BROKEN =1 X 436 kJ/mol

TOTAL = 4359kJ/mol

TWO C-C BONDS ARE FORMED = 2 X 347 kJ/mol

SEVEN C-H BONDS ARE FORMED = 7 X 414 kJ/mol

ONE C-O BOND FORMED = 1 X 351kJ/mol

ONE O-H BOND IS FORMED = 1 X 464 kJ/mol

TOTAL =4407kJ/mol

ENTHALPY CHANGE = 4359 - 4407

= -48 kJ/mol

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.