Consider the reaction: CO2(g)+CCl4(g) ⇌2 COCl2(g) Calculate ΔG for this reaction at 25^∘ C under

Consider the reaction: CO2(g)+CCl4(g) ⇌2 COCl2(g) Calculate...

Consider the reaction: $$\mathrm{CO}_{2}(g)+\mathrm{CCl}_{4}(g) \rightleftharpoons 2 \mathrm{COCl}_{2}(g)$$ Calculate $\Delta G$ for this reaction at $25^{\circ} \mathrm{C}$ under the following conditions: $$ \begin{aligned} &P_{\mathrm{CO}_{2}}=0.112 \mathrm{~atm} \\ &P_{\mathrm{CCl}_{4}}=0.174 \mathrm{~atm} \\ &P_{\mathrm{COCl}_{2}}=0.744 \mathrm{~atm} \end{aligned} $$

Consider the reaction:
$$\mathrm{CO}_{2}(g)+\mathrm{CCl}_{4}(g) \rightleftharpoons 2 \mathrm{COCl}_{2}(g)$$
Calculate $\Delta G$ for this reaction at $25^{\circ} \mathrm{C}$ under the following conditions:
$$
\begin{aligned}
&P_{\mathrm{CO}_{2}}=0.112 \mathrm{~atm} \\
&P_{\mathrm{CCl}_{4}}=0.174 \mathrm{~atm} \\
&P_{\mathrm{COCl}_{2}}=0.744 \mathrm{~atm}
\end{aligned}
$$

Key Concepts

Ideal Gas Approximation

The ideal gas approximation assumes that gases behave ideally, meaning their molecules have negligible interactions and occupy no volume. This assumption allows the use of partial pressures directly as activities in the calculation of the reaction quotient, simplifying the determination of ?G for gaseous reactions.

Standard State Conditions

Standard state conditions provide a reference point for thermodynamic values, typically defined as 1 atm for gases and 1 M for solutions. The standard Gibbs free energy change (?G°) corresponds to the free energy change when all reactants and products are in their standard states, allowing the adjustment to actual conditions via the term RT ln Q.

Reaction Quotient

The reaction quotient (Q) is a dimensionless number calculated from the partial pressures or concentrations of the reactants and products. For gaseous reactions, Q is determined using the partial pressures raised to the power of their stoichiometric coefficients, and it describes the system's status relative to equilibrium.

Gibbs Free Energy

Gibbs free energy (?G) is a thermodynamic quantity that indicates the amount of work obtainable from a chemical reaction at constant temperature and pressure. It is used to predict the spontaneity of a reaction, where a negative ?G signifies a spontaneous process under the given conditions.

Transcript

00:01 In this question we've been given a reaction in which co2 reacts with ccl4 reversibly to produce 2 moles of co cl2 and we've been asked to determine the overall gives energy change of this reaction and by definition we know that this is equal to the standard gives energy change plus rt lim q say lim qp this link this qp this is just an equilibrium constant of partial pressures that is the products the ratio of the partial pressures of the products to that of the reactants so we have the partial pressure of co cl2 over the partial pressures of co co to end pc cl4 and remember we have to raise these to their respective co -efficients in the overall reaction this is true so we raise this to the power of 2 now looking at the we might as well just start by calculating this this being equal to 0 .744 over 0 .112 times 0 .174 and this is square so our qp this is going to be equal to 28 .404.

01:36 Now going back to the standard gives energy change.

01:40 Here we are looking at the summation of the standard gives energy change of formation of the products minus that of the reactants of the reactions and we always have to make sure we multiply by the respective stoichiometric coefficients and if we are to look at this system this is our product and these are our reactants.

02:09 So we are saying this minus the reactants.

02:12 And if we get our values, we'll retrieve our values from our appendices.

02:17 This is going to be the standard gives energy change is going to be 2 by negative 204.

02:28 .9 minus one more multiplied by negative 394 .4 minus 1 more multiplied by negative 394 .4 minus 1 more multiplied by negative 394 .4, minus 1 more, multiplied by negative 6 to 0 .3 so our standard gives energy change is going to be equal to 4 6 .9 and this is in kilojohns now if we are to use this formula we then plug these values that would have calculated we now have calculated our standard gives energy change and we've calculated our qp so our temperature has to be in kelvin so it's going to be 298 .15 kelvin.

03:09 So the gibbs energy change of that reaction is 46 .9, multiplied by 10 to the power 3...

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