Question

A) For a particular process that is carried out at constant pressure, q = 145 kJ and w = -35 kJ. Therefore,

= 180 kJ and ΔH = 145 kJ.

ΔE = 145 kJ and ΔH = 180 kJ.

ΔE = 110 kJ and ΔH = 145 kJ.

ΔE = 145 kJ and ΔH = 110 kJ. B) Dry chemical hand warmers utilize the oxidation of iron to form iron oxide according to the following reaction: 4Fe(s)+3O2(g)→2Fe2O3(s) Standard thermodynamic quantities for selected substances at 25 ∘C Reactant or product ΔH∘f(kJ/mol) Fe(s) 0.0 O2(g) 0.0 Fe2O3(s) −824.2

Calculate ΔH∘rxn for this reaction.

Express the energy in kilojoules to one decimal place.

C)

Calculate ΔHrxn for the following reaction:

C(s)+H2O(g)→CO(g)+H2(g)

Use the following reactions and given ΔH values:

C(s)+O2(g)→CO2(g), ΔH= -393.5 kJ
2CO(g)+O2(g)→2CO2(g), ΔH= -566.0 kJ
2H2(g)+O2(g)→2H2O(g), ΔH= -483.6 kJ

Express your answer using four significant figures.

ΔHrxn =

.... kJ

D) How much heat must be absorbed by a 23.0 g sample of water to raise its temperature from 30.0 ∘C to 50.0 ∘C? (For water, Cs=4.18J/g∘C.)

Express your answer to three significant figures and include the appropriate units.

Answer 1

1)

q = 145 KJ

w = -35 KJ

use:

ΔE = q + w

= 145 KJ - 35 KJ

= 110 KJ

ΔH = q at constant pressure

= 145 KJ

Answer: 110 KJ , 145 KJ

2)

Given:

Hof(Fe(s)) = 0.0 KJ/mol

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

Hof(Fe2O3(s)) = -824.2 KJ/mol

Balanced chemical equation is:

4 Fe(s) + 3 O2(g) ---> 2 Fe2O3(s)

ΔHo rxn = 2*Hof(Fe2O3(s)) - 4*Hof( Fe(s)) - 3*Hof(O2(g))

ΔHo rxn = 2*(-824.2) - 4*(0.0) - 3*(0.0)

ΔHo rxn = -1648.4 KJ

Answer: -1648.4 KJ

3)

Lets number the reaction as 0, 1, 2, 3 from top to bottom

required reaction should be written in terms of other reaction

This is Hess Law

required reaction can be written as:

reaction 0 = +1 * (reaction 1) -0.5 * (reaction 2) -0.5 * (reaction 3)

So, ΔHo rxn for required reaction will be:

ΔHo rxn = +1 * ΔHo rxn(reaction 1) -0.5 * ΔHo rxn(reaction 2) -0.5 * ΔHo rxn(reaction 3)

= +1 * (-393.5) -0.5 * (-566.0) -0.5 * (-483.6)

= 131.3 KJ

Answer: 131.3 KJ

4)

Given:

m = 23 g

C = 4.184 J/g.oC

Ti = 30 oC

Tf = 50 oC

use:

Q = m*C*(Tf-Ti)

Q = 23.0*4.184*(50.0-30.0)

Q = 1924.6 J

Answer: 1920 J

After rounding to three significant figures