Heat reaction question?
I'm doing this chapter on heat reactions, enthalpies, etc. and I am absolutely confused. I've banged my head against the wall 21 times. I just don't understand it. Here's a question that is a good example of how all my other questions are. Could you not just give the answer, but show me how to do it? Thanks. What is the final temperature, in degrees C, when 60.0 g of water at 80 degrees C is mixed with 40.0 g of water at 25 degrees C. The specific heat of water is 4.184 J/g x C. a) 53 b) 58 c) 70 d) 35 e) 42 Thanks in advance
Chemistry question?
So for the reaction we're given you end up with 200.0 mL with a density of 1.00g/mL so you have 200.0g with the heat capacity of 4.184 J/gC. To raise the 200g 12.1 degrees you will need (200g)(12.1C)(4.184J/gC) = -10125.28J (negative because energy is released) Molar Weight of Ca = 40.08g/mol moles of Ca = 0.745g/(40.08 g/mol) = 0.0186 mol Ca moles of HCL = .2L(.5 mol/L) = 0.1 mol HCl Ca is the limiting reactant. If 0.0186 mol Ca generates -10125.28J, then 1 mol generates -10125.28J/0.0186 mol = -544kJ/mol
Chemistry heat of reaction/Hess's Law help?
The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy charges: 1) NO(g) + NO2(g) -----> N2O3(g) deltaH = -39.8 KJ 2) NO(g) + NO2(g) + O2 -----> N2O5 (g) deltaH = -112.5 KJ 3) 2NO2(g) -----> N2O4(g) deltaH = -57.2 KJ 4) 2NO(g) + O2(g) ------> 2NO2(g) deltaH = -114.2 KJ 5) N2O5(s) ------> N2O5(g) deltaH = 54.1 KJ Calculate the heat of reaction for: N2O3(g) + N2O5(s) ------> 2N2O4(g) I know this would be calculated using Hess's Law, but I for some reason cannot figure it out! Any help would be greatly appreciated!! :)
Heat of Reaction Delta H Chemistry Question!?
For the first question, you should use formula ∆H= ∑Bonds Broken - ∑Bonds Formed Remember for any reaction,the bonds in the reactants are broken and in the products bonds are formed. 1.You have to use the values of bond energies given in your text book or syllabus. the required bond energies are listed below. all values are given in kj/mol a.O-H 460 b.C-C 350 c.C-H 410 d.O=O 496 e.C=O 740 2.now calculation part,according to the balanced chemical equation ∑Bonds Broken=2[6(C-H)+2(C-C)] + 9[O=O] = 2[2460+700] +9[740] =12980kj/mol when taking C3H6 make its structure it is alkane with 6CH bonds and 2CC bonds to be broken.i have just substituted the values of bond energies.all other cases are also same. similarly, ∑Bonds Formed= 6[2(C=O)] + 6[2(O-H) = 6[1480] + 6[920] = 14400kj/mol 2. now,according to the above mentioned formula, ∆H= ∑Bonds Broken - ∑Bonds Formed = 12980-14400 =(-)1420kj/mol negative sign showing the reaction is exothermic. 2.Now next part of the question. mass of C3H6=2.16 g mole of C3H6= mass/mr (Mr. of C3H6=42) =0.0514moles therefore, 1mole of C3H6 releases 1420kj energy(amount of heat) 0.0514moles will release 0.0514*1420=73.0kJ of energy. Hope, this works
Heat of reaction, conceptal questions?
I waited to answer this but as no one else has I will give my opinion.It has been a while and I am not that familiar with the equation and what the variables are. Having said that what I do know and I hope it helps is; 1) When you dissolve NaOH in water there is heat liberated, it is an exothermic reaction. The amount of heat liberated is due to the Na+ and OH- separating into ions. As such the amount of energy liberated is based on the amount of NaOH added. The increase in temperature of the solution will depend on the amount of the solution. ie add 10g of NaOH to 1 liter of water and a certain amount of energy will be liberated and there will be a corresponding increase in the temperature of the water. Add the same amount of NaOH to 2 litres of water and the same amount of energy will be released but as there is more solution to absorb the energy the increase in temperature will be correspondingly lower. 2)Same thing as 1 except that it is a chemical reaction that is releasing the energy. Again it is the combination of HCl and NaOH that will be releasing the energy. But there is a catch. You need to know the limiting reagent. As only the chemicals that participate in the reaction will release energy. eg NaOH and HCl react on a molar basis. 1 mole of NaOH will react with 1 mole of HCl. So if you have 1 mole of NaOH and 1 mole of HCl the will react completely to release a certain amount of energy. Now take the situation were everything else is the same except that you have 2 moles of HCl and 1 mole of NaOH. As only 1 mole of HCl can participate in the reaction the other mole of HCl will do nothing so the amount of energy released is the same. I hope this helps. If you are still stuck send me an email and I will look up my text books to try to give you an exact answer.
General Chemistry heat of reaction problem?
ok first there are 175 mL of total reaction volume - that is also 175 g for the reaction mass using 1g/mL conversion. Next the heat rise of the solution is 2.9 C which is 2.9 K. For this experiment we ignore any heat loss to the outside world.;) 0.125 L of 0.25 M CsOH solution is 0.03125 moles CsOH Now... 2.9K x 175 g x 4.2 J/g-K = 2131.5 J for this reaction. 2131.5 J/0.03125 moles = 68208 JOULES/mole that would be 68.2 KJ/mol Did you forgot to change J to KJ? 68000 KJ/mol is a big amount of energy!
Chemistry: heat of formation?
First, let's write the reaction: CaCO3 -> CO2 + CaO We know the heat of the overall reaction, as well as the heats of formation of two of the three components. Thus, we can calculate the heat of formation of the third. (Sum of Heats of Formation of Products) - (Sum of Heats of Formation of Reactants) = Heat of Reaction Substituting in what we know: (-393.5 kJ/mol + X) - (-1206 kJ/mol) = 176.9 kJ/mol Solving for X: X = 176.9 - 1206 + 393.5 = -635.6 kJ/mol That would be the heat of formation of the calcium oxide, or lime.
Chemistry problem regarding the heat of reaction for the combustion of Titanium?
this is much simpler than using that formula. It looks complicated, but its actually just trying to throw you off using Kelvin degrees... yet each increase in kelvin degree is exactly the same as an increase in a Celcius degree. Thus, the increase from 25C to 53.8C is 28.8 If each degree increase requires 9.84 kJ the total energy increase is 28.8x9.84 = 283.392kJ This is for 0.721grams of Titanium You need to find the kJ increase per mole. Seeing that titanium has an atomic mass of 47.9, the moles of 0.721grams of Titanium is: 0.721/47.9 = approximately 0.015moles Thus per 0.015051 moles, the kJ increase is 283.392kJ dividing by 0.015051, per 1mole, the increase is 283.392kJ / 0.015052 which gives you 18,827.3kJ per mole Using standard form and rounding up to 3 significant figures, this is the same as 18.88x10^4 kJ/mol heat given out. Watch out, though! This is energy given OUT, an exothermic reaction, therefore the heat of combustion is negative, ie -1.88 x 10^4 kJ/mol
What are the chemical reactions of copper and heat?
If heated in open air to a certain temperature, copper would undergo the reaction below.[math] 2Cu(s) + O_2(g) → 2CuO(s) [/math]CuO is a black solid so you would expect your copper turn black.
Heat of Formation from Heat of Reaction?
∆H˚rxn = ∑∆H˚f(products) - ∑∆H˚f(reactants) ∆H˚rxn = 2 ∆H˚f(CuO (s)) - [ ∆H˚f(Cu2O (s)) + 1/2 ∆H˚f(O2 (g))] but, ∆H˚f (O2 (g)) = 0 (by definition, for any element in its most stable form, the value is always zero. -146 kJ = 2∆H˚f(CuO (s)) - (-168.6 kJ) -146 kJ - 168.6 kJ = 2∆H˚f(CuO (s)) -314.6 kJ/2 = ∆H˚f(CuO (s)) -157.3 kJ = ∆H˚f(CuO (s))