Is the reaction spontaneous at 298 K? Work: Delta G= -76kJ/mol - 298K(-117J/K/1000) Delta G= -41.134 Spontaneous 2) A reaction has a delta H of 11 kJ and a delta S of 49 J/K. c) In which direction is the reaction, as written, spontaneous at 25 Cġ) A reactions has a delta H of -76 Kj and a delta S of -117 J/K. Delta H rxn = -633.1 kJ/mol a) Calculate the value of Delta S rxn at 25.0 C. C3H8(g) + 2O2(g) =>įor the reaction described by the chemical equation: 3C2H2(g) -> C6H6(l). Then indicate if the reaction is entropy driven, enthalpy driven or neither. For each system below indicate whether DELTA S and DELTA H are positive or negative. 2)Either Delta is anĢCH3CH2CH2OH(l)+9O2(g)→6CO2(g)+8H2O(g) I have to find the delta G delta G of CH3CH2CH2OH(l)=−360.5 KJ/Mol delta G of O2(g)=0 KJ/Mol delta G of CO2(g)=−394.4 KJ/Mol delta G of H20(g)=−228.6 KJ/Mol I keep getting -3426.2 KJ/Mol but its not correct.Ĭalculate the DELTA H(rxn), DELTA S(rxn), DELTA S(universe), DELTA G(rxn). Either Delta is an alpha or Delta is a theta. If Delta is a theta, then Delta is a beta. 1) If Delta is an alpha, then Delta is a beta.
#Dfind dy and delta y how to
I really dk how to do them,i have spend hours doing it Can somebody please solve and if possible show the steps thx,just 3 questions _ 3)Given that tan theta = -✓3 and that 90 degree ≤ delta ≤ 180 degree,find the value of: a) tan delta b)cos deltaĭetermine which statement follows validly from the statements included. Given the thermodynamic data in the table below, calculate the equilibrium constant for the reaction: 2SO2(g)+O2-> 2SO3 Substance (DeltaH^o) (Delat S^o) SO2 -297 249 O2 0 205 SO3 -395 256 Answer (it was given) 2.32x10^24 Even though the answer is given BUT THAT DOES NOT MAKE SENSE BECAUSE THE CHANGE IN ENTHALPY OF THE ORIGINAL DELTA H1 ISĬonsider the following reaction at 298K 4Al(s)+3O2(g) yields 2Al2O3(s) delta H= -3351.4 Kj find delta S sys find delta S surr find delta S unvi I know I posted this question before, but can you clarify it? MY QUESTION IS AFTER YOU SWITCHED THE DELTA H1 THE CHANGE IN THE ENTHALPY IS NEGATIVE. Calculate the temperature at which Keq is 4.0*10^3. Arrange the following solutions in order ofįor the reaction I2(s)+Cl2(g) => 2ICl(g) delta H= 36kJ, delta S= 158.8J/K at 25° C. k is a constant and need not enter into the calculations. i is the number of particles i.e., Na3PO4 will have i = 4 (3 for Na and 1 for PO4). Deltaĭelta T = ikfm delta T is the amount f.p. The enthalpy changes for two different hydrogenation reactions of C2H2 are: C2H2+H2->C2H4 Delta H 1 C2H2+2H2->C2H6 Delta H 2 Which expression represents the enthalpy change for the reaction below? C2H4+H2->C2H6 Delta H = ? A. Okay, and then the difference between the actual distance between them should be 1.25.Using differentials find dy and delta y when y=7/x and dx=delta x =1 So it's crossing the x axis at zero and 4 and it's got a its vertex is halfway in between there. Um So you have this parabola X squared minus four X. Okay then it says draw a picture, like a picture somewhere else, I don't know. So if you didn't have to get really close then then that's good enough for approximation probably. The real change and our approximation is one. 5, which is why? And So we got Delta Y is 1.25. So you want to find it at three? How much was it at? Three? So two times three minus four which is six minus four, which is to and so D Y is two times. So we gotta find the derivative of that function F. Can the formula from the Y is the derivative at X times dx Okay. Y is the approximate change in case it was too hard to find your calculator or whatever approximate change in why as exchanges From 3 to 3.5. Why would move And would there would be a difference of 1.25 between them? Why would move from -3 -1.75.
![dfind dy and delta y dfind dy and delta y](https://d2vlcm61l7u1fs.cloudfront.net/media/cfc/cfca3e20-8509-425d-affb-8a1093a88ba8/php3rAF8o.png)
So if we looked at a picture of this as we moved from X equals 32 X equals 3.5. So then the delta Y is Why of 3.5 -Y of three Which is -1.75 -3. 3.5 sq to a point 25 minus four times 3.5 14. And then why at 3.5? And that would be 3.5 square calculator. So, I want you to first figure out why at three And that would be 9 -12 which is -3.
![dfind dy and delta y dfind dy and delta y](https://i.stack.imgur.com/C6mu9.jpg)
It means the actual real change in why as exchanges from 3- 3.5.
![dfind dy and delta y dfind dy and delta y](https://i.ytimg.com/vi/4DS3q2Ao2FM/maxresdefault.jpg)
And that and get this delta Y and the wife. You're giving this function? Why? It's X squared minus four X.