7) . Each of the terms on the right side of the equation are either endothermic

Question

7). Each of the terms on the right side of the equation are either endothermic or exothermic. Which answer properly depicts this.

Endothermic, Exothermic, Endothermic

Endothermic, Endothermic, Endothermic

Exothermic, Exothermic, Endothermic

Exothermic, Endothermic, Endothermic

Endothermic, Endothermic, Exothermic

8)Find the ?Hrxn for the reaction: 3C(s)+4H2(g) ? C3H8(g)
Using these reactions with known ?H's:

C3H8(g)+5O2(g)?3CO2(g)+4H2O(g)   ?H=?2043 kJ

C(s)+O2(g)?CO2(g)   ?H=?393.5 kJ

2H2(g)+O2(g)?2H2O(g)   ?H=?483.6 kJ

Express your answer in kJ.

9)A 12.0 g sample of a metal is heated to 90.0 C. It is then dropped into 25.0 g of water. The temperature of the water rises from 22.5 to 25.0 C. The specific heat of water is 4.18 Jg-1C-1. Calculated the specific heat of the metal. Express your answer in Jg-1C-1. Enter only a numerical value, do not enter units.

10)What is the process by which a solute forms a solution in a solvent?

dissolution

bumping

delineation

neutralization

calisthenics

Explanation / Answer

(8) The given thermochemical reactions are

C3H8(g)+5O2(g) --à 3CO2(g)+4H2O(g)   DeltaH1= - 2043 kJ ------(1)

C(s)+O2(g)--à CO2(g)   DeltaH2 = - 393.5 kJ ------(2)

2H2(g)+O2(g)--à 2H2O(g)   DeltaH3= - 483.6 kJ -----(3)

We need to calculate DeltaH(rxn) for the following chemical reaction

3C(s)+4H2(g) --à C3H8(g)   -------(4)

In the above equation (4), we need 3C(s) and 4H2(g) on left.

Hence we need to multiply eqn(2) by 3 and eqn(3) by 2 and add them to get all the reactants.

Also we need one C3H8(g) on product side, hence we need to reverse eqn(1) and add them to get eqn(4). Hence

(2)x3=> 3C(s)+3O2(g)--à 3CO2(g)   DeltaH5 = - 1180.5 kJ ------(5)

(3)x2=> 4H2(g)+2O2(g)--à4H2O(g)   DeltaH6= - 967.2 kJ -----(6)

Rev(1)=>3CO2(g)+4H2O(g)-à C3H8(g)+5O2(g), DeltaH7= -(- 2043 kJ) = +2043 kJ ----(7)

Now adding eqn(5), (6) and (7), we get eqn(4) i.e

3C(s)+4H2(g) --à C3H8(g)   -------(4)

Hence DeltaH(rxn) = DeltaH5+ DeltaH6+ DeltaH7

= (- 1180.5 kJ)+( - 967.2 kJ)+2043 kJ = - 104.7 KJ (answer)

(10) The process by which a solute forms solution in solvent is called dissolution.

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

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