Calculate ΔG^∘ for H2 O(g)+(1)/(2) O2(g) ⇌H2 O2(g) at 600 . K using the following data: H2( g)+O

step 1 There are multiple steps involved in this question. To solve for delta G for the chemical reacti

Calculate ΔG^∘ for H2 O(g)+(1)/(2) O2(g) ⇌H2 O2(g) at 600 ....

Calculate $\Delta G^{\circ}$ for $\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)$ at $600 . \mathrm{K}$ using the following data: $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6}$ at $600 . \mathrm{K}$ $2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37}$ at 600. $\mathrm{K}$

Calculate $\Delta G^{\circ}$ for $\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)$ at $600 . \mathrm{K}$
using the following data:
$\mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6}$ at $600 . \mathrm{K}$
$2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37}$ at 600. $\mathrm{K}$

Key Concepts

Temperature Effects in Thermodynamic Calculations

Temperature plays a critical role in determining the value of Gibbs free energy through its appearance in the equation ?G° = -RT ln K. The factor RT shows that even modest changes in temperature can significantly affect the free energy change of a reaction, thereby influencing its spontaneity and equilibrium position.

Reaction Manipulation and Combining Equilibria

When reactions are algebraically manipulated—such as adding, subtracting, or multiplying them by a scalar—the overall equilibrium constant for the resulting reaction is determined by specific rules. For instance, when reactions are added, the overall equilibrium constant is the product of the individual constants, and when a reaction is reversed or its stoichiometric coefficients are scaled, the equilibrium constant is inverted or raised to a corresponding power. This principle allows one to derive the equilibrium constant for a target reaction from known reactions.

Equilibrium Constant

The equilibrium constant (K) quantifies the relative concentrations or activities of reactants and products at equilibrium. It reflects the extent to which a reaction proceeds and is temperature-dependent. A larger K indicates that the products are favored at equilibrium, while a smaller K implies that the reactants are predominant.

Gibbs Free Energy

Gibbs free energy (?G°) is a thermodynamic quantity that indicates the spontaneity of a reaction. It is calculated using the relation ?G° = -RT ln K, where R is the universal gas constant, T is the temperature in Kelvin, and K is the equilibrium constant. This equation links the chemical equilibrium of a reaction with its energetic favorability under standard conditions.

Transcript

00:01 There are multiple steps involved in this question.

00:05 To solve for delta g for the chemical reaction, given two chemical reactions and their equilibrium constants, we first need to use hess's law to figure out how we can combine those two chemical reactions to get the overall reaction.

00:21 Then, after combining those two reactions, we need to look at their equilibrium constants and combine, use those equilibrium constants.

00:31 To determine the equilibrium constant of the overall reaction.

00:36 Then, once we have the equilibrium constant of the overall reaction, we can calculate delta g for the overall reaction.

00:43 The overall reaction is h2o plus 1 .5 .02 goes to h202 at 600 degrees cell, or 600 kelvin.

00:52 We want to calculate delta g.

00:54 I don't want to give it all away here at once.

00:57 So the two chemical reactions for which we have information are h2, plus o2 goes to h202, with an equilibrium constant of 2 .3 times 10 to the 6.

01:10 Then we have 2h2 plus o2 goes to 2h2 with an equilibrium constant of 1 .8 times 10 to the 37.

01:18 In order to combine these two chemical reactions to get this overall reaction, we need to reverse the second reaction and divide by 2, or multiplied by 1 half.

01:29 When we do that, we get this reaction here, and what is its new equilibrium constant? well, if we divide a reaction by two or multiply it by one -half, we need to raise this k -value to one -half.

01:45 Also, if we reverse the chemical reaction, we need to take the reciprocal of the k -value, so we need to do both.

01:53 The new k -value will be one over the original k -value raised to the one -half.

02:02 We'll then cancel what's common to both sides, and we do get the overall reaction up above...

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