1. 4Al( s ) + 3O 2 ( g ) 2Al 2 O 3 ( s ) A.1.986 × 10 3 kJ B.73.61 kJ C.147.2 kJ

Question

1.

4Al(s) + 3O2(g) 2Al2O3(s)

A.1.986 × 103 kJ

B.73.61 kJ

C.147.2 kJ

D.36.80 kJ

E.18.40 kJ

2.

Which of the following statements is true concerning the decomposition of liquid water to form hydrogen gas and oxygen gas?

2H2O(l) 2H2(g) + O2(g)

A.H is greater than U because of the pressure–volume work done by the gaseous products.

B.H equals U because both are state functions.

C.H is greater than U because the pressure is constant.

D.H is less than U because of the pressure–volume work done by the gaseous products.

E.H is less than U because the atmosphere does pressure–volume work on the gaseous products.

3.

The molar mass of an unknown gas was measured by an effusion experiment. It was found that it took 54 s for the gas to effuse, whereas nitrogen gas required 48 s. The molar mass of the gas is:

A.32 g/mol

B.26 g/mol

C.30 g/mol

D.35 g/mol

E.22 g/mol

Explanation / Answer

Answer –

1) We are given, mass of Al = 1.185 g

4Al(s) + 3O2(g) ------> 2Al2O3 (s)

Hf of Al2O3 = -1676 kJ/mol

Horxn = sum of Hoproduct - sum Ho reactant

             = [2* HoAl2O3(s)] – [4 Ho Al(s) + 3 HoO2(g) ]

            = (2*-1676 kJ/mol) – ( 4*0.00 + 3*0.00 )

           = -3352 kJ

Now we need calculate the moles of Al

Moles of Al = 1.185 g / 26.982 g.mol-1

                    = 0.0439 moles

From the reaction

4 moles of Al = -3352 kJ

So, 0.0439 moles of Al = ?

= 0.0439 moles of Ak * -3352 kJ / 4 moles of Al

= -36.8 kJ

So answer is D.36.80 kJ

2) D.H is less than U because of the pressure–volume work done by the gaseous products

We know the relation between H and U

U = H + pV

Means as the change in volume increase the value for U increase. We are given 2 moles of reactant and it is in the liquid phase and it is converted to 3 moles of gases means there is volume increase so and pressure also increase, so there is work done by the gaseous product , so answer is D.

3) We are given, rate of effusion of unknown gas = 54 s

Rate of effusion of N2 gas =48 s , we know molar mass of N2 = 28.014 g/mol

We know Graham's law of effusion

Rate1 / Rate 2 = M2/ M1

54 s / 48 s = 28.014 g.mol-1 / M1

1.125 = 28.014 g.mol-1 / M1

Taking square from both side

(1.125)2 = 28.014 g.mol-1 / M1

So, M1 = 28.014 / (1.125)2

             = 22.13 g/mol

So answer is E.22 g/mol

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