Question

1) Acetylene (C2H2) is used in welding torches. How much heat is produced if 6.55gC2H2 is combusted with excess oxygen gas?

C2H2(g)+52O2(g)⟶2CO2(g)+H2O(l)ΔH∘c=−1301.1kJ/molqsystem<0 and wsystem>0

a. qsystem>0 and wsystem>0

b. qsystem>0 and wsystem<0

c. qsystem<0 and wsystem<0

2) Acetylene (C2H2) is used in welding torches. How much heat is produced if 6.55gC2H2 is combusted with excess oxygen gas?

C2H2(g)+52O2(g)⟶2CO2(g)+H2O(l)ΔH∘c=−1301.1kJ/mol

Answers

a. −327kJ

b. −8520kJ

c. −0.131kJ

d. −419kJ

3) What is the standard enthalpy of formation (at 1atm and 25∘C) for O2(g)?

Answers

a. 0kJ/mol

b. 25kJ/mol

c. 100kJ/mol

d. −50kJ/mol

4)

When 2.16g of H2 reacts with excess O2 by the following equation, 627kJ of heat are produced. What is the enthalpy change of the reaction, per mole of hydrogen gas?

2H2+O2⟶2H2O

Answers

a. −41.8kJ

b. −271kJ

c. −585kJ

d. −671kJ

Answer 1

2)

Given

Mass of acetylene = 6.55 g

Molar mass of aacetylene = 26 g/mol

No. of moles of acetylene = mass / molar mass = 6.55 g / 26 g/mol = 0.251 mole

Given

H = -1301 KJ/mol

q = H * no. of moles = -1301 KJ/mol * 0.251 moles = -327 KJ Answer (a) - 327 KJ

3) heat of formation an element will always be zero O2 is an element

so Answer is (a) 0 KJ/mol

4) 2.16 g of H2

Molar mass of H2 = 2 g/mol

No. of moles = 2.16 g / 2 g/mol = 1.08 mol

given q = 627 KJ

H = q /n = 627 KJ/1.08 mol = 585 Kj/mol

Answer is (c) - 585 KJ/mol