Calculate the delta H rxn of the following reaction? Need to figure this out by tonight please!?
I don't even know where to start with this question. Somebody help please? If you can help with either one that'd be great. Thanks in advance! 1. Calculate DeltaHrxn for the following reaction: Fe2O3(s) + 3CO(g) --- 2Fe(s) + 3CO2(g) Use the following reactions and given Delta Hs 2Fe(s) + 3/2O2 --- Fe2O3 (s), Delta H = -824.2 Kj CO(g) + 1/2 O2(g) ---- CO2(g), Delta H = -282,7 2. Calculate DeltaHrxn for the following reaction: Use the following reactions and given delta H's: C(s) + 2H2 (g) --- Ch4 (g) Delta H = -74.6 kj C(s) + 2Cl2 (g) --- CCl4 (g) Delta H = -95.7 kj H2(g) + Cl2 (g) --- 2HCl (g) Delta H = -184.6 kj
Calculate the standard enthalpy of reaction (deltaHrxn) for the following reaction using the data below.?
2 ClF3 = +338 kJ 2 NH3 = +92 kJ 6 HF = - 1626 kJ +338 kJ +92 kJ - 1626 kJ = -1196 kJ So: 2 ClF3(g) + 2 NH3(g) → N2 + 6 HF(g) + Cl2(g), ΔH = -1196 kJ
Calculate the standard enthalpy change for the reaction 2a+b⇌2c+2d use the following data:?
1. Use the fact that the sum of dH*stoich coeff for the products minus that of the reactants will give you the dH of raction. so (2*(-497)+(2*185)) - (2*(-267)+1*(-387)) = -624 kJ/mol + 921 kJ/mol = 297 kJ/mol so this reaction is endothermic (energy absorbing), if 3.70 mol A reacts then, 1.85 mol of B reacts, 3.70 mol of C forms 3.70 mol of D forms. So take (3.70 mol A)*(-267 kJ/mol) = -987.9 kJ (1.85 mol B* -387 kJ/mol) = -715.95 kJ (3.70 mol C * 185 kJ/mol) = 684.5 kJ (3.70 mol D * -497 kJ/mol) = -1838.9 kJ Now take products minus reactants (C+D) - (A+B) = -1154.4 kJ/mol +1702.9 kJ/mol = 548.5 kJ of heat is absorbed. This is such a funky reaction.
Help calculating Delta H rxn using standard enthalpies of formation?
This Site Might Help You. RE: Help calculating Delta H rxn using standard enthalpies of formation? Propane (C3H8) burns according to the following balanced equation: C3H8(g) + 5O2(g)--->3CO2(g) + 4H2O(g) Calculate Delta H Rxn for this reaction using standard enthalpies of formation. (The standard enthalpy of formation of gaseous propane is -103.9 kJ/mol .)
Calculations -Standard Enthalpy of formation?
ΔH°reaction = sum of standard enthalpy of formation of products - sum of standard enthalpy of formation of reactants ΔH°reaction = [ 2 X ΔH°f for H2O + 2 X ΔH°f for SO2 ] - [ 2 X ΔH°f for H2S + 3 X ΔH°f for O2 ] ΔH°reaction = [ 2 X -285.8 + 2 X -296.8 ] - [ 2 X -20.63 + 3 X 0 ] ΔH°reaction = [-571.6 - 593.6 ] - (-41.26) ΔH°reaction = -1165.2 + 41.26 = -1123.94 kj elements in their standard state will have zero heat of formation so for O2 g ΔH°f is zero .... see the standard enthalpy of formation of these compounds at this link-- http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_(data_table)
Calculate delta H for the following reaction?
Okay, this is related to Hess' Law which states that the enthalpy change of a reaction is independent of the path taken. The standard enthalpy of formation of a compound is the heat energy released when one mole of the compound is formed from its constituent elements in their standard physical states under standard conditions. You could imagine the reactants on the left hand side of an equation being deconstructed into their constituent elements in their standard physical states(heat energy is absorbed when this happens) and the product(s) on the right hand side of that equation being formed from their constituent elements(heat energy is released when this happens). Having visualised this, it would make sense that we subtract the sum of the standard enthalpies of formation of the reactants from those of the products. (1) ΔHf(C₅H₁₂)-5ΔHf(CO₂)-6ΔHf(H₂O)= -3505.8kJ (2) ΔHf(CO₂)= 393.5kJ --->ΔHf(C) and ΔHf(O₂) are 0kJ since carbon and oxygen are in their standard physical states (3) ΔHf(H₂O)= 285.83 -->I believe that there's an error in the information provided since it is the ΔHf of liquid water that is needed, not that of gaseous water. I checked the ΔHf values of all the reactants and it appears that the values only tally if the ΔHf of liquid water is used. ΔHf(C₅H₁₂)-5(393.5kJ/mol)-6(285.83kJ/m..... -3505.8kJ ΔHf(C₅H₁₂)= -3505.8kJ+5(393.5kJ/mol)+6(285.83kJ/mol) = 176.68kJ/mol Since C and H₂ are in their standard physical states, ΔHrxn = -ΔHf(C₅H₁₂) = -176.68kJ I hope this helps and feel free to send me an e-mail if you have any doubts!
You usually calculate the enthalpy change of combustion from enthalpies of formation.The standard enthalpy of combustion is [math]ΔH_c^∘[/math].It is the heat evolved when 1 mol of a substance burns completely in oxygen at standard conditions. For example,[math]C_{2}H_{2}(g)+\dfrac{5}{2}O_{2}(g)→2CO_{2}(g)+H_{2}O(l)[/math]You calculate [math]ΔH_c^∘ [/math]from standard enthalpies of formation:[math]ΔH_c^∘=∑ΔH_f^∘(p)−∑ΔH_f^∘(r)[/math]where, [math]p[/math]stands for "products" and [math]r [/math]stands for "reactants".For each product, you multiply its [math]ΔH_f^∘ [/math]by its coefficient in the balanced equation and add them together.Do the same for the reactants. Subtract the reactant sum from the product sum.EXAMPLE:Use the following enthalpies of formation to calculate the standard enthalpy of combustion of acetylene, [math]C_{2}H_{2}[/math].[math]C_{2}H_{2}(g)+\dfrac{5}{2}O_{2}(g)→2CO_{2}(g)+H_{2}O(l)[/math][math]C_{2}H_{2}(g)=226.73 kJ/mol ; [/math][math]{ΔH_{CO_2}^o}(g)=-393.5 kJ/mol ;[/math][math]{ΔH_{H_2{O}}^o}(l)=-285.8 kJ/mol[/math]Solution:[math]C_{2}H_{2}(g)+\dfrac{5}{2}O_{2}(g)→2CO_{2}(g)+H_{2}O(l)[/math][math]ΔH_c^∘=∑ΔH_f^∘(p)−∑ΔH_f^∘(r)[/math][math][2 × (-393.5) + (-295.8)] – [226.7 + 0] kJ=-1082.8 - 226.7= -1309.5 kJ[/math]The heat of combustion of acetylene is [math]-1309.5 kJ/mol.[/math]
You need to find the enthalpy of formation of the reactants and the products, and take the difference. This gives you the enthalpy change at the reference temperature of your data. To get it at any other temperature you need to calculate the enthalpy input to get the reactants to that temperature, and add it to the enthalpy of formation. (Molar specific heat x temperature difference x amount). Do the same for the product and again take the difference.For other reactants make sure to allow for the stoichiometry.
Delta H reaction for the following reaction: C(s)+H2O(g)=>CO(g)+H2(g) Use the following reactions a?
Reverse the second and third equations. After doing so change the signs of the Delta H's of said equations. After that half the same equations. C(s)+O2(g)==>CO2(g) Delta H = -393.5kJ 2CO(g)+O2(g)=>2CO2(g) Delta H= -566.0 kJ 2H2(g)+O2(g)=>2H2O(g) Delta H= -483.6 kJ then becomes: C(s)+O2(g)==>CO2(g) Delta H = -393.5kJ CO2(g)==>CO(g)+.5O2(g) Delta H = 283.0 kJ H2O(g)==>H2(g)+.5O2(g) Delta H = 241.8 kJ After cancelling like terms you will end up with the target equation, C(s)+H2O(g)==>CO(g)+H2(g). Add the Delta Hs and you'll get 131.3 kJ
Steps for calculation of enthalpy of wet steam are the following:From the steam tables note the specific enthalpy, hs of dry saturated steam needs to be noted.The actual enthalpy of evaporated wet steam is the product of the dryness fraction, ζ and the specific enthalpy, hs from the steam tables.Wet steam has less usable heat energy than dry saturated steam.ht = hs ζ + (1 - ζ) hw (1)whereht = enthalpy of wet steam (kJ/kg, Btu/lb)hs = enthalpy of ateam (kJ/kg, Btu/lb)ζ = dryness fractionhw = enthalpy of saturated water or condensate (kJ/kg, Btu/lb)The dryness fraction can be expressed asζ = ws / (ww + ws) (2)whereζ = dryness fractionww = mass of water (kg, lb)ws = mass of steam (kg, lb)