Ethene (C2 H4) can be halogenated by the reaction: C2 H4(g)+X2(g) ⇌C2 H4 X2(g) where X2 can be C

step 1 Here we're going to calculate delta H, delta S, delta G and K. For this equation, replacing the

Ethene (C2 H4) can be halogenated by the reaction: C2...

Ethene $\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)$ can be halogenated by the reaction: $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{X}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{X}_{2}(g)$$ where $\mathrm{X}_{2}$ can be $\mathrm{Cl}_{2}, \mathrm{Br}_{2},$ or $\mathrm{I}_{2} .$ Use the thermodynamic data given to calculate $\Delta H^{\circ}, \Delta S^{\circ}, \Delta G^{\circ},$ and $K_{\mathrm{p}}$ for the halogenation reaction by each of the three halogens at $25^{\circ} \mathrm{C}$ . Which reaction is most spontaneous? Least spontaneous? What is the main factor responsible for the difference in the spontaneity of the three reactions? Does higher temperature make the reactions more spontaneous or less spontaneous?

Ethene $\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)$ can be halogenated by the reaction:
$$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{X}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{X}_{2}(g)$$
where $\mathrm{X}_{2}$ can be $\mathrm{Cl}_{2}, \mathrm{Br}_{2},$ or $\mathrm{I}_{2} .$ Use the thermodynamic data given to calculate $\Delta H^{\circ}, \Delta S^{\circ}, \Delta G^{\circ},$ and $K_{\mathrm{p}}$ for the halogenation reaction by each of the three halogens at $25^{\circ} \mathrm{C}$ . Which reaction is most spontaneous? Least spontaneous? What is the main factor responsible for the difference in the spontaneity of the three reactions? Does higher temperature make the reactions more spontaneous or less spontaneous?

Key Concepts

Thermodynamic Quantities

In chemical thermodynamics, key quantities such as the standard enthalpy change (?H°), the standard entropy change (?S°), and the standard Gibbs free energy change (?G°) are used to characterize reactions. ?H° indicates whether a reaction is exothermic or endothermic, ?S° reflects the degree of disorder or randomness change during the reaction, and ?G° combines these factors to predict the overall spontaneity of a reaction at a given temperature.

Standard Gibbs Free Energy Change (?G°) and Reaction Spontaneity

?G° is a crucial metric in determining if a reaction will proceed spontaneously under standard conditions. A negative ?G° means that a reaction is thermodynamically spontaneous, while a positive value indicates non-spontaneity. The relation ?G° = ?H° - T?S° shows how both the reaction’s enthalpy and entropy contribute to the free energy change, hence affecting spontaneity.

Equilibrium Constant (Kp) and its Relationship to ?G°

The equilibrium constant for a reaction (Kp), which is expressed in terms of pressures for gaseous reactions, is directly related to the standard Gibbs free energy change by the expression ?G° = -RT ln Kp. This relation connects the thermodynamic predictions of spontaneity with the extent of reaction under equilibrium, providing insight into the reaction’s favorability.

Bond Energies and Reaction Enthalpy

The differences in the spontaneity of similar reactions can often be traced to variations in bond strengths, which influence the reaction enthalpy (?H°). Generally, reactions involving the formation or breaking of stronger bonds will have different enthalpic profiles, which is a key factor in determining the overall energy change and hence the reaction's spontaneity.

Temperature Effects on Reaction Spontaneity

Temperature plays a significant role in reaction spontaneity by affecting the T?S° term in the Gibbs free energy equation. An increase in temperature can enhance or reduce spontaneity depending on whether the reaction is driven by enthalpy or entropy. Thus, analyzing the sign and magnitude of ?S° helps in predicting if a higher temperature will make a reaction more spontaneous or less spontaneous.

Transcript

00:02 Here we're going to calculate delta h, delta s, delta g and k.

00:07 For this equation, replacing the x with cl2, i2, and br2.

00:13 I've already given the delta h and s values from the appendix.

00:18 So let's start with the equation.

00:20 Let's start with cl2.

00:23 So that makes our equation c2h4 plus cl2 gives c2h4, gives c2h4.

00:34 Cl2.

00:36 So we're going to start with calculating delta h using the equation delta h is equal to summation of delta 8 of the products minus summation of n delta h of the reactants.

00:52 So we already know the product here is c2h for cl2.

00:58 Delta h for that is minus 129 .7.

01:07 A reactants are c2h4 .50.

01:14 2 .4 plus delta h for c l2 is 0 so this gives us a value of minus 1 82 .1 kilojou next we're going to calculate delta s using the same reaction using the same formula delta s is equal to summation of n delta s of the products minus summation of n delta s of the reactants so this gives us a value of or we know the products as c2h for co2 308 minus c2h4 -19 .3 plus 223 .1.

02:19 So this gives us a final delta s value of minus 134.

02:25 0 .4 .4 j .5 .com.

02:32 And now to calculate delta g, we use the formula delta g is equal to delta h minus t times delta s.

02:43 We know that the delta h value is minus 182 .1 kilojoules minus 298 times minus 134 .4 .4 jol per kelvin, 298 kelvin.

03:06 But this is in jules and delta age is in kilojols.

03:11 So i'm writing this into kilojoules.

03:17 This gives us a final delta g value of minus 142 kilojoules.

03:24 And now to calculate delta g, to calculate the k value, you do delta g is equal to minus r t, l and k.

03:40 So we know all the values here except k.

03:43 So we put in minus 142 kilojoules.

03:52 When r is in jolts, and we can work this by thousand to give us jolts, minus 8 .314 into 298 into lm .k.

04:12 Solving for k, we get 7 .78 into 10 to the power of 24.

04:20 So these are the values of delta h, delta s, delta g, and k for this equation with cl2.

04:40 Now moving on to br, br2, our new equation becomes c2h4 plus br2...

Please give Ace some feedback