B. for the following questions, use the reaction no2(g) n2(g) + o2(g), with δh = –33.1 kj/mol and δs= 63.02 j/(mol·k).

i. draw a possible potential energy diagram of the reaction. label the enthalpy of the reaction.

ii. is the reaction endothermic or exothermic? explain your answer. (2 points)

iii. what is the gibbs free energy of the reaction at 25°c?

iv. is the reaction spontaneous or nonspontaneous at 25°c? explain your answer.

I. Kindly, see the attached image.

II. The reaction is exothermic.

III. - 51.88 kJ/mol.

IV. The reaction is spontaneous.

Explanation:

I. Draw a possible potential energy diagram of the reaction. Label the enthalpy of the reaction.

In an exothermic reaction, the energy of the reactants is higher than that of the products.

Kindly see the attached image to show you the potential energy diagram of the reaction.

II. Is the reaction endothermic or exothermic? Explain your answer.

If the sign is positive, the reaction is endothermic.

If the sign is negative, the reaction is exothermic.

Herein, ΔH = - 33.1 kJ/mol, so the reaction is exothermic.

III. What is the Gibbs free energy of the reaction at 25°C?

∵ ΔG = ΔH - TΔS.

Where, ΔG is the Gibbs free energy change (J/mol).

ΔH is the enthalpy change (ΔH = - 33.1 kJ/mol).

T is the temperature (T = 25°C + 273 = 298 K).

ΔS is the entorpy change (ΔS = 63.02 J/mol.K = 0.06302 J/mol.K).

∴ ΔG = ΔH - TΔS = (- 33.1 kJ/mol) - (298 K)(0.06302 J/mol.K) = - 51.88 kJ/mol.

IV. Is the reaction spontaneous or nonspontaneous at 25°C?

The sign of ΔG indicates the spontaneity of the reaction:

If ΔG < 0, the reaction is spontaneous.

If ΔG = 0, the reaction is at equilibrium.

If ΔG > 0, the reaction is nonspontaneous.

Herein, ΔG = - 51.88 kJ/mol, so the reaction is spontaneous.

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.