loading

The perfect choice of one-stop service for diversification of architecture.

2021-12-09

Digah Company

29

**Gibbs free energy and Kp help!!?**

G sys = H sys - TS sys T is in kelvin You can find delta H and delta S from table at standard condition. H sys= 2(H H2) H o2 - 2( H H2o) In the same way find S sys After you find G find Kp from this equation.....Kp = exp(-G/RT) where R is gas constant=8.314 and T is in kelvin Regards

â€” â€” â€” â€” â€” â€”

**How do I find enthalpy change from standard free energy change in this scenario?**

G=H-TS I think you will have to play around a little bit to find h though. do it for both scenarios and set h=h. good luck

â€” â€” â€” â€” â€” â€”

**Why do people think the concept of free energy is impossible?**

Zeus forbid that solar panel technology should become economically feasible to install on the roof of every house in America. Which it would, if the big boys were not shy of not holding the public hostage to their Power Bills. No, it's better to pay General Electric to build huge, inefficient windmills. Power monopolies are for the People, not for the Money. I bow to their eternal, wind blown wisdom

â€” â€” â€” â€” â€” â€”

**Why hasn't Tesla's 'Radient Energy' discovery led to eneryone having free energy everywhere?**

The government would not make any money off that. That is the only reason

â€” â€” â€” â€” â€” â€”

**Why isn't standard Gibbs free energy always zero?**

$Delta G = Delta G^circ - RT ln(K)$It is not directly $G$, but formula for change in free energy. Also, in standard conditions $Delta G = Delta G^circ$, so it is not necessarily 0.Relevant reading on a website

â€” â€” â€” â€” â€” â€”

**Gibbs free energy or reaction enthalphy**

If you measure the heat absorbed or released by the process at constant volume, you are measuring $Delta U$ (the change in internal energy). You will need to measure the temperature change of the reaction when it's running in a rigid, sealed container---or, better, compute it from $Delta H$ measured as follows.If you measure the heat absorbed or released by the process at constant pressure, you are measuring $Delta H$ (the change in enthalpy). You can measure the temperature change of the reaction when it's running in a container open to the atmosphere, e. g. a coffee cup calorimeter.If you are measuring the nonexpansion work reversibly and at constant temperature and pressure, you are measuring $Delta G$ (the change in Gibbs free energy). You can do your measurements in a container open to the atmosphere immersed in a constant temperature bath. The system must be closed in the sense that you are not losing any reactants or products

â€” â€” â€” â€” â€” â€”

**is free energy a threat to america's future.?**

>it seems we spend an awful lot of time wasted.

â€” â€” â€” â€” â€” â€”

**After all this time debating about GW, why can't people recognize that free energy devices will partly solve..?**

Sure, energy is free. Study physics. Free energy is right up there with perpetual motion in efficiency.

â€” â€” â€” â€” â€” â€”

**Calculate the standard free-energy change at 25 Celsius for the following reaction:?**

I am on mastering chemistry and it requires the answers to be in joules making it -3.71*10^5 J

â€” â€” â€” â€” â€” â€”

**What is your favorite non-sugar free energy drink?**

Rooster Booster. It's the best. I love the taste, it's not nasty and it's like a pop it has carbination in it. I usually grab one after the gym at the gas station, it's in a fountain and it's like .75 cents not like 3.

â€” â€” â€” â€” â€” â€”

**Is the Gibbs free energy not/less important for canonical ensembles? If so, why?**

$G$ and $F$ are the thermodynamic potentials of different ensembles, the one for $G$ has, as far as I know, no commonly accepted name. The one for $F$ is the canonical ensemble. The bath in the ensemble corresponding to $G$ can exchange volume and energy with the system and is characterized by two intense parameters the system gets into equilibrium with: temperature and pressure. For a system characterized by the common parameters $S$, $N$, $V$ there are 8 ensembles depending on which are considered fixed or exchanged with the bath. The bath is then characterized by the corresponding intensive parameters ($T$, $mu$ and $p$).The thermodynamic potentials that are extremal in the equilibrium of those ensembles are related by Legendre transformations that change the variables from a quantity to the derivative of the function with respect to that quantity (yes, the same Legendre transformation as in Hamiltonian mechanics). The potential of the microcanoical ensemble is the energy $E(S, V, N)$ expressed in terms of the natural variables. From there we get to the Free energy by a Legendre transform $F = E - TS$, where $T = partial_S E$ and $S$ must be eliminated by calculating and substituting $S(T)$ to express $F$ in terms of its so called natural variables.As a general thermodynamic result the potential corresponding to the ensemble with the bath parameters $(T, mu, p)$ is zero. This gives the Gibbs-Duhem relation: $$ S , dT - V,dp N,dmu = 0 $$If we introduce other quantities describing the overall system (e.g. magnetization) there are even more ensembles and we again can change between extensive and intensive parameters by Legendre transformation.

GET IN TOUCH WITH Us