Question

What will the final temperature of 400 mL sample of water at 39.2°C be if 36,555 J of heat were added.

a.88.1°C

b.58.2°C

c.77.0°C

d.61.0°C

How much heat is produced when 6.26g of zinc metal reacts with oxygen gas under standard conditions? Use standard enthalpies of formation provided in your textbook.

a. 35.5 kj

b.267 kj

c.1021 kj

68.7 kj

Ethanol, C2H5OH, is used as a primary fuel for motor vehicles in some countries like brazil,The balanced chemical equation for the combustion reaction is

C2H5OH(l) + O2(g)-------- CO2(g) + H2O(g)

The density of the ethanol is 0.7893g/mL .Determine the enthalpy of combustion of exactly 2L of ethanol.

a. -3.31 x 10^4 kj

b.-4.68 x 104

c.-1.11 x 104 kj

d.1.69 x 10^4 kj

Answer 1

Answer

(I)-

Given:-

volume of water (V) = 400 mL

Heat energy (q) = 36555 J

Initial temperature (Tinitial) = 39.2 0C

Final temperature (Tfinal) = ?

Since we know that

density of water (d) = 1.0 g / mL

specific heat of water (C) = 4.18 J / g.0C

As we know that

mass (m) = volume(V) density(d)

So

mass of water (m) = volume of water (V) density of water (d)

mass of water (m) = 400.0 mL 1.0 g / mL

mass of water (m) = 400.0 g

According to formula

Heat energy (q) = mass of water (m) specific heat of water (C) change in temperature (T)

Heat energy (q) = mass of water (m) specific heat of water (C) change in temperature[Final temperature (Tfinal) - Initial temperature (Tinitial) ]

36555 J = 400.0 g 4.18 J / g.0C (Tfinal  - 39.2 0C)

36555 J = 1672.0 J /  0C (Tfinal  - 39.2 0C)

Tfinal  - 39.2 0C = 36555 J / 1672.0 J /  0C

Tfinal  - 39.2 0C = 36555 J / 1672.0 J /  0C

Tfinal  - 39.2 0C = 21.86 0C

Tfinal = 21.86 0C + 39.2 0C

Tfinal = 61.0 0C

therefore correct option 'd' i.e 61.0 0C (i.e the answer).

(II)-

Given:-

wt. of zinc (Zn) = 6.26 g

Since we know that

molar mass of Zn = 65.38 g / mol

molar mass of O2 gas = 2 molar mass of O

molar mass of O2 gas = 2 16

molar mass of O2 gas = 32 g / mol

molar mass of ZnO = molar mass of Zn + molar mass of O

molar mass of ZnO = 65.38 + 16

molar mass of ZnO = 81.38 g /mol

Also we know that

standard enthalpy of formation of ZnO(s) i.e (H0ZnO) = - 348.0 kJ/mol

standard enthalpy of formation of Zn(s)  i.e (H0Zn) = 0.0 kJ/mol

standard enthalpy of formation of O2(g)  i.e (H0O2) = 0.0 kJ/mol

So

2Zn(s) +  O2(g)       2 ZnO(s)

therefore

standard enthalpy change of reaction(H0rxn) = H0products - H0reactants

standard enthalpy change of reaction(H0rxn) = [2 standard enthalpy of formation of ZnO(s) i.e (H0ZnO)] - [ 2 standard enthalpy of formation of Zn(s) i.e (H0Zn) + standard enthalpy of formation of O2(g)  i.e (H0O2) ]

standard enthalpy change of reaction(H0rxn) = (2 - 348.0 kJ/mol ) - ( 2 0.0 kJ/mol + 0.0 kJ/mol )

standard enthalpy change of reaction(H0rxn) = - 696.0 kJ/mol - 0.0 kJ/mol

standard enthalpy change of reaction(H0rxn) = - 696.0 kJ/mol

So

2 Zn(s) + O2(g)       2 ZnO(s) ; H0rxn = - 696.0 kJ/mol

2 65.38 g 32 g   2 81.38 g

130.76 g 32 g 162.76 g

therefore

130.76 g Zn reacts with O2 gas produced = - 696.0 kJ heat energy

1.0 g Zn reacts with O2 gas produced = - 696.0 / 130.76 kJ heat energy

then

6.26 g Zn reacts with O2 gas produced = - 696.0 6.26 / 130.76 kJ heat energy

6.26 g Zn reacts with O2​​​​​​​ gas produced = - 4356.96‬  / 130.76 kJ heat energy

6.26 g Zn reacts with O2​​​​​​​ gas produced = - 33.2 kJ heat energy

Above value of produced heat energy i.e 33.2 kJ is equivalent to the 35.5 kJ therefore correct option is 'a; i.e 35.5 kJ (i.e the answer)

(II)-

Given:-

C2H5OH(l) + O2(g) CO2(g) + H2O(g)

density of ethanol (C2H5OH) = 0.7893 g / mL

volume of of ethanol (C2H5OH) = 2.0 L

Since

1.0 L = 1000 mL

So

volume of of ethanol (C2H5OH) = 2.0 L = 2.0 1000 mL = 2000 mL

As we know that

mass of ethanol (C2H5OH) = volume of of ethanol (C2H5OH) density of ethanol (C2H5OH)

mass of ethanol (C2H5OH) = 2000 mL 0.7893 g / mL

mass of ethanol (C2H5OH) = 1578.6 g

Also we know that

molar mass of ethanol (C2H5OH) = 2 molar mass of C + 5molar mass of H + molar mass of O + molar mass of H

molar mass of ethanol (C2H5OH) = 2 12 + 5​​​​​​​1 + 16 + 1

molar mass of ethanol (C2H5OH) = 24 + 5 + 16 + 1

molar mass of ethanol (C2H5OH) = 46 g /mol

similarly

molar mass of O2 = 32 g /mol

molar mass of CO2 = 44 g/mol

molar mass of H2O = 18 g /mol

Since we know that combustion of ethanol (C2H5OH) is as follows:-

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g) ; H0combustion = - 1366.95 kJ / mol

46 g 96 g 88 g 54

As we know that

The enthalpy of combustion of 46 g ethanol (C2H5OH) =   - 1366.95 kJ

The enthalpy of combustion of 1.0 g ethanol (C2H5OH) =   - 1366.95 / 46 kJ

then

The enthalpy of combustion of 1578.6 g ethanol (C2H5OH) =  - 1366.95 1578.6 / 46 kJ

The enthalpy of combustion of 1578.6 g ethanol (C2H5OH) =  - 2157867.27 / 46 kJ

The enthalpy of combustion of 1578.6 g ethanol (C2H5OH) =  - 46910.15 kJ

or

The enthalpy of combustion of 1578.6 g ethanol (C2H5OH) =  - 4.69 104 kJ

As mentioned above value of enthalpy of combustion i.e - 4.69 104 kJ which is equivalent to the - 4.68 x 104 kJ therefore correct option is 'b' i.e 4.68 x 104 kJ (answer)