Question

Use the ΔH°f values provided to determine ΔH°rxn for the following reaction CH4(g) + 3 Cl2(g) → CHCl3(l) + 3 HCl(g) ΔH°rxn = ? ΔH°f (kJ/mol): -75 -134 -92 A. +662 kJ B. +117 kJ C. -151 kJ D. -335 kJ The equation that corresponds to the enthalpy of formation for NH3(g) is A. N(g) + 3H(g) → NH3(g) B. N(g) + 3/2 H2(g) → NH3(g) C. 1/2 N2(g) + 3H(g) → NH3(g) D. 1/2 N2(g) + 3/2 H2(g) → NH3(g) 17. Given the first thermochemical equation, determine ∆H for the second one. 4 AlCl3(s) + 3 O2(g) → 2 Al2O3(s) + 6 Cl2(g) ∆H = –529.0 kJ Cl2(g) + ⅓ Al2O3(s) → ⅔ AlCl3(s) + ½ O2(g) ∆H = ? A. +529.0 kJ B. +88.2 kJ C. +176.3 kJ D. +264.5 kJ explain please

Answer 1

1)

Given:

Hof(CH4(g)) = -75.0 KJ/mol

Hof(Cl2(g)) = 0.0 KJ/mol

Hof(CHCl3(l)) = -134.0 KJ/mol

Hof(HCl(g)) = -92.0 KJ/mol

Balanced chemical equation is:

CH4(g) + 3 Cl2(g) ---> CHCl3(l) + 3 HCl(g)

ΔHo rxn = 1*Hof(CHCl3(l)) + 3*Hof(HCl(g)) - 1*Hof( CH4(g)) - 3*Hof(Cl2(g))

ΔHo rxn = 1*(-134.0) + 3*(-92.0) - 1*(-75.0) - 3*(0.0)

ΔHo rxn = -335 KJ

Answer: -335 KJ

Anaswer: D

2)

for delta Horxn to be delta Hof,

rule 1: there must be only 1 mol of product being formed

rule 2: reactant must be element in there standard state

A. N(g) + 3H(g) → NH3(g)

Standard state of Hydrogen is H2(g)

So, this is wrong

B. N(g) + 3/2 H2(g) → NH3(g)

Standard state of Nitrogen is N2(g)

So, this is wrong

C. 1/2 N2(g) + 3H(g) → NH3(g)

Standard state of Hydrogen is H2(g)

So, this is wrong

D. 1/2 N2(g) + 3/2 H2(g) → NH3(g)

This is correct

Answer: D

Only 1 question at a time please