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2 H2(g) + O2(g) →2 H2O (g) + 130 kJ calculate Gibbs free energyat standard temperature if the value for entropy was fond to be833JChemistry 1405...

Question

2 H2(g) + O2(g) →2 H2O (g) + 130 kJ calculate Gibbs free energyat standard temperature if the value for entropy was fond to be833JChemistry 1405

2 H2(g) + O2(g) →2 H2O (g) + 130 kJ calculate Gibbs free energy at standard temperature if the value for entropy was fond to be 833J Chemistry 1405



Answers

The Gibbs free energy of formation, $\Delta_{f} G^{\circ}$, of HI is $+1.70 \mathrm{~kJ} / \mathrm{mol}$ at $25^{\circ} \mathrm{C}$. Calculate the equilibrium constant for the reaction $\mathrm{HI}(\mathrm{g}) \Longrightarrow \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{~J}_{2}(\mathrm{~g})$.

Hello students In this question we have to calculate the entropy change in the surrounding that is delta is surrounding. When one mole of number of moles are one of hydrogen that is water in the liquid form is formed under dutch tender condition and that and healthy of formation It is equal to -286 kg per more. Okay, so we can right here that the it is given that this is the end healthy of formation for the even bowl. So hence the heat absorbed Q. By the surrounding it will be equals two plus 2 86.3 to 86. Hello, june Per mold. Okay. Or we can say that this is the absorbed energy by one bowl. So this is the heat absorbed by the one mode. So we can calculate that the changing and property for the surrounding it is equal to heat absorbed, divided by temperature T. So heat absorbed is to 86. Close jewel develop the temperature teeth Which is room temperature, so which can be taken as to 98 Calvin. So from here this is in the colonial, so can be converted into jewels, manipulate and to the power plus three. So changing and property for the surrounding it is equals to the 9 59.7 three jewels per mole kelvin. So this becomes the answer for this problem. Okay, thank you.

In this question, we will be covering the topic of Gibbs free energy and understanding how to calculate the energy change for this reaction. Now for the reaction here I have the Delta G values listed for each of the reactant and products. Now, if you notice here for age too, it does not have a value. Ah, the reason being is if you remember Di atomic molecules or molecules such as H 202 and two F two. Those molecules don't have a Delta G value because they're in their most stable form, meaning that there is no reason for them to change, hence having no Delta G value. So let's begin to answer the question or begin to solve the equation here. The equation here is Delta G is equal to Delta G of the products minus Delta G of the reactions, so we would typically just plug in those values. Delta G. So for the products here we have, don't forget the stoic geometric coefficient. Um, so you would write it as this two, since we have to molecules of it or Mel's. So it's two times the value is negative. Nine six nine. So that's minus the reactant. So for the reactant, since it is 87 now there's six moles of water. So we would we basically have to some these two up, So 87 plus it's going to be six times negative. You draw that better to 37 So after we calculate that Delta G is equal to negative 19 38 minus and all these in the parentheses here equal to negative 13 three, five. So, in conclusion, Delta G is equal to negative. 603. And yes, all these values were given and killer jewels Permal. And this is your final answer.

So far are given a chemical reaction. That is two K two in the solid state generates K 20 and 3/2 2. So we have our Delta G the reaction equation because we want to calculate Gibbs Energy. Remember that oxygen and Standard State we have it's free energy is equal to zero. So why Delta G After we consider the other components, there's 159.9 killed Jules Month. So next we're calculating K so we can use the following equation So the G not is equal to negative r t l N k So we can rearrange for lnk. What we get is negative 64.208 and then we simply just So if okay, that is 1.30 times 10 to the negative 28. So k two is the most dynamically stable with respect to cater. I want to

So we're continuing our work with thermodynamics. Really? In the context off our main group chemistry. We're looking at the Gibbs Energy. The change in an therapy, the change in entropy. And we're gonna be looking at how that Inter relates to our chemical equations that we are able to form. So the chemical equation of interest we have here is to okay Oh, to in the solid state generates K 20 solid at three of the two 02 in the gay c'est state. So we're looking for the free energy of reaction on this is define er's Dr G react. So we have these some delta G react off product minus product minus our reactors. So therefore we have a value of 159 point 09 kg jewels pommel. So what I've done here is I've used the values that we have been provided with. I haven't calculated these myself, but I've just made sure that I've taken into account my stroke geometric values. So what we know is that Delta G not equals negative. R t Alan Kay. We then three a range Well, Alan Kay Andi solve. Okay, Well, what we're doing by solving for K is we're incorporating an exponential into our equation. Okay, Is exponential negative? 64 0.208 on, then we solved. Okay, 1.30 times 10 the minus 28. So, co two is Thumma dynamically more stable than Cato, l and 02 at 298 Calvin.


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