D Question 7 10 pts Calculate the standard enthalpy change for the following rea

Question

D Question 7 10 pts Calculate the standard enthalpy change for the following reaction given that the AHf of benzene, C6H6 is 49 kJ/mol, the AHCf of CO2(g) 394, /mol and AHof of H20 242 kJ/mol 2C6H6(l) 1502 12CO2(g) 6H20(g) AH O -6082 kJ O +6082 O 6278 O 6278 O -7013 kJ D Question 8 10 pts The specific heats of Lead and Gold are both is 0.128 J g OC. Which of the following statements is not true O Lead has the higher molar heat capacity O Gold and Lead have the same molar heat capacity O When 10.0 g of lead is heated from 25°C to 35°C, it gains128 J of heat O When 10.0 g of gold is heated from 25°C to 35°C, it gains128 J of heat O None of the statements are false

Explanation / Answer

7)

Given:

Hof(C6H6(l)) = 49.0 KJ/mol

Hof(O2(g)) = 0.0 KJ/mol

Hof(CO2(g)) = -394.0 KJ/mol

Hof(H2O(g)) = -242.0 KJ/mol

Balanced chemical equation is:

2 C6H6(l) + 15 O2(g) ---> 12 CO2(g) + 6 H2O(g)

Ho rxn = 12*Hof(CO2(g)) + 6*Hof(H2O(g)) - 2*Hof( C6H6(l)) - 15*Hof(O2(g))

Ho rxn = 12*-394.0 + 6*-242.0 - 2*49.0 - 15*0.0

Ho rxn = -6278 KJ

Answer: -6278 KJ

I am allowed to answer only 1 question at a time

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

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