Question

Part 1:

Which of the following gas samples would be most likely to behave ideally under the stated conditions? Explain.

A) N₂ at 1 atm and -70°C

B) CO at 200 atm and 25°C

C) SO₂ at 2 atm and 50 °C

D) Ne at STP

E) NH₃ at 1 atm and 25°C

PART 2:  Given that

                      Na(l) + ½ O₂ (g) ⇌   NaO (g)                              K1 = 2 x 10⁵

                      Na₂O₂(s)   +   Na(l) ⇌ NaO(g) + Na₂O(s)      K2 = 2 x 10 ^{̶ 8}

                      2Na(l) + 1/2O₂ (g)      ⇌ Na₂O(s)                    K3 = 2 x 10²⁵

                Determine the value for the equilibrium constant for the following reaction

                                         Na₂O₂(s)    ⇌   2NaO(g)

Answer 1

part 1

Correct answer is D) Ne at STP

Explanation: Real gas behaves as ideal when intermolecular forces of attraction in them is almost zero. This usually occurs at high temperature and low pressure.

A) Here temperature is -70°C, i.e. very low. So,ideal behaviour is not possible.

B) Here pressure is 200 atm i.e. very high. So, ideal behaviour is not possible

C) Here temperature as well as pressure is more. So, ideal behaviour is not possible.

D).Here temperature and pressure are of STP and Ne molecules do not have forces of intermolecular attraction.

E) Here conditions are favorable but NH3 forms hydrogen bonds which increases intermolecular atttractions. So, it cant be ideal gas.

part 2

K1 = [NaO]/[O2]1/2

Pure substances such as liquid sodium are not included in expression for equilibrium constant. Also concentration of pure solids and pure liquids is considered as unity. These are avoided in expression for K.

K2 = [NaO]

K3 = 1/[O2]1/2

K of Desired reaction: [NaO]2 = (K2)^2

= (2×10-8)2

= 4 x 10-16. (Answer)