Question

 

1. Consider the following thermochemical equation: 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g) ∆H° = –878.2 kJ

(a) How much heat is released when 3.0 mol ZnS(s) reacts in excess oxygen?

(b) How much heat is released when 2.3 × 10^-2 mol ZnS(s) reacts in excess oxygen?

(c) What is the enthalpy change when 223.9 g ZnS(s) reacts in excess oxygen?

(d) What is the enthalpy change when 0.96 g ZnO(s) is produced?

2. Slaked lime (Ca(OH)2(s)) is produced when lime (calcium oxide, CaO(s)) reacts with liquid water. 65.2 kJ of heat is released for each mol of Ca(OH)2 that is produced.

(a) Write a thermochemical equation for the reaction.

(b) What is the enthalpy change when 523.3 kg of lime reacts with excess water?

3. The following reaction represents the complete combustion of hexane, C6H14(l) + 19/2O2(g) → 6CO2(g) + 7H2O(l) ∆H°= –4163 kJ

(a) If 0.537 mol of carbon dioxide is produced in the reaction represented by the equation above, how much heat is released by the reaction?

(b) If 25.0 kg of hexane is burned in sufficient oxygen, how much heat will be released?

(c) What mass of hexane is required to produce 1.0 × 10⁵ kJ of heat by complete combustion?

Answer 1

1a)given :2ZnS(s)+3O₂(g)--->2ZnO(s)+2SO₂(g)

delta H=-878. 2KJ

a)Heat released for 3 mol of ZnS =?

b)Heat released for 2.3*10^-2mol=?

C)Enthalpy change for 0.96g ZnO produced=?

Solution :

From balanced chemical equation, we can see that 2 moles of Zns reacts with 3 moles of oxygen to form moles of zinc oxide and 2 moles of sulfur dioxide. The heat released =-878. 2KJ

Thus,we can set two equations as follows :

2 mol of ZnS releases heat = - 878.2KJ

3mol of ZnS releases heat =

- 878.2KJ*3mol/2mol=2634. 6/2KJ =1317. 3KJ

b)2 mol of ZnS releases heat = -878. 2KJ

2.3*10^-2mol of ZnS releases heat

= - 878.2KJ*2.3*10^-2mol/2mol =-20.1986/2KJ=-10.0993KJ

After round off, enthalpy change =10.1KJ

C) calculate mole of ZnS

Mole = given mass /molar mass =223.9g/97.45g/mol=2.298mol

From equation

2 mol of ZnS produces heat = - 878.2KJ

2.298 mol of ZnS produces heat = - 878.2KJ *2.298mol/2mol=2018.1036/2KJ=1009.0518KJ

Enthalpy change =1009.05KJ

(molar mass =Atomic mass of Zn+Atomic mass of S =65.38g/mol +32.07g/mol=97.45g/mol)

D)step 1: Convert grams of ZnO into moles

Molar mass of ZnO = Atomic mass of Zn +Atomic mass of O =65.38g/mol+16g/mol=81.38g/mol

Moles of ZnO =mass of ZnO /molar mass of ZnO =0.96g/81.38g/mol=0.01178mol

From equation we can see that 2 moles of ZnO are formed. THUS, we can write as follows :

2 mol of ZnO leads to enthalpy change=-878. 2KJ

0.01178mol of ZnO leads to enthalpy change =-878. 2KJ*0.01178mol/2mol=10.345196/2KJ

=5.172598KJ

After round off, enthalpy change for ZnO =-5. 2KJ

2)BALANCED CHEMICAL EQUATION

CaO +H₂O - - - - >Ca(OH) ₂ delta H = - 65.2KJ

Given:mass of lime =523.3Kg

Convert into gram

1Kg =1000g

523.3Kg =1000g*523.3Kg/1Kg =523.3*10³g

Here, 1 mol of CaO releases heat = - 65.2KJ

1mol of CaO = molar mass of CaO=56.08g

(molar mass of CaO =Atomic mass of Ca +Atomic mass of O =40.08g/mol+16.00g/mol =56.08g/mol)

56.08g of CaO releases heat = - 65.2KJ

523.3*10³g of CaO releases heat = - 65.2KJ*523.3*10³g/56.08g =34119160/56.08KJ =608401.5692 KJ

After round off, Enthalpy change =608.401*10³

ALTERNATIVE METHOD

Step 1: Convert gram into moles

Moles = grams of lime / molar mass of lime =523.3*10³g/56.08g/mol=9.3313*10³mol

Step 2: set two equations

1 mol of CaO releases heat =-65. 2KJ

9.3313*10³mol of CaO releases heat

= - 65.2KJ *9.3313*10³mol/1mol= 608.401*10³KJ

3)a)In given equation,

6 mol of CO ₂ releases heat =-4163KJ

0.537mol of CO₂ releases heat =

- 4163KJ*0.537mol/6mol = -2235.531/6KJ=-372.5885KJ

b) given : 25Kg of hexane

Convert it in grams

1 kg =1000g

25Kg =25000g

1 mole of hexane(C₆H₁₄ )= molar mass =86.2g

(Molar mass = 6*Atomic Mass of C+14*Atomic mass of H =6*12.01g/mol+14*1.01g/mol =86.2g/mol)

86.2g of hexane releases heat =-4163KJ

25000g of hexane releases heat = - 4163KJ*25000g/86.2g =104075000/86.2KJ=1207366.589KJ

Or heat released =1207.37*10³KJ

C) From equation, we can see that 1mole of hexane =molar mass =86.2g produces heat =-4163KJ

-4163KJ of heat is produced by combustion of 86.2g of hexane.

1.0*10⁵KJ of heat is produced by combustion of 86.2g*1.0*10⁵KJ/-4163KJ =86.2*10⁵/-4163KJ

=-0.02071*10⁵g

Thus,heat produced =2.071*10³g

Or 2.1*10³g(Two significant figures)