Question

Graphite solid is placed in a closed chamber of fixed volume with an initial oxygen pressure of 200 atm and heated to T=1300°C. Assume the graphite is not completely consumed during the subsequent reaction. (a) If the total pressure is held constant, what ratio of carbon dioxide to carbon monoxide is produced by any reactions that occur? (b) If the reaction is allowed to proceed from the initial conditions without allowing any gas in or out, would the chamber pressure increase, decrease, or remain the same? (c) If the chamber is perfectly insulated, would the chamber temperature increase, decrease, or remain the same? (d) How – if at all – would your answers to the previous parts change if all the graphite is consumed?

C_graphite + 1/2O_2(g) à CO_(g) ∆G₁° = -111,700 – 87.65 T (J/mol)

C_graphite + O_2(g) à CO_2(g) ∆G₂° = -394,100 – 0.84 T (J/mol)

Answer 1

Part- (a.) - the given reactions are:

Cgraphite + 1/2 O₂(g) CO(g)

Suppose the moles of oxygen consumed in this reaction is 'n'.

Therefore, 'n' moles of oxygen produces '2n' moles of CO.

similarly, for second equation is:

Cgraphite + O₂(g) CO2(g)

'n' moles of oxygen will produce 'n' moles of CO₂ .

Since other factors like pressure volume temmperature being constant.

Therefore, Ratio of ratio of carbon dioxide to carbon monoxide is produced = n/2n = 1:2

Part- (b.) - Since reaction is happening in an close container(constant volume) at constant temperature, therefore change in pressure will depend upon change in number of moles.

As the product is forming, by stoichiometry , 1 mole of O₂ , produces 2 moles of CO. Means, the number of moles in the product increase. As a result, the pressure of the container also increases as the reaction proceeds.

Part- (c.)-    If the chamber is perfectly insulated, the temperature of the chamber increases, because according to the standard Gibb's Free Energy(given) ,

∆G₁° = -111,700 – 87.65 T (J/mol)

∆G₂° = -394,100 – 0.84 T (J/mol)

These equations can be compared to a standard equation,  ∆G° = ∆H - T∆S,

This shows ∆H(Change in Enthalpy) for both reactions is "Negative", which implies reaction is an "Exothermic Reaction".

Hence the temperature of the chamber increases.

Part- (d.) - The answers of previous parts remains same, if all the graphite is consumed or not.