How do you calculate the standard enthalpy of formation of acetylene from the heat of combustion of C2H2,C (graphite) and H2 given as -1300kj/mol, 395kj/mol and -286kj/mol respectively?
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]
Acetylene, C2H2, can be converted to ethane, C2H6, by a process known as hydrogenation. The reaction is?
C2H2(g) + 2H2(g) <----> C2H6(g) Delta Gf for this reaction is: Delta Gf = (1)(209.2) - (1)(-32.89) - (2)(0) = 242.09 kJ/mol Remember: Delta Gf for reaction = sum of moles of each product times delta Gf of each product, minus the sum of moles of each reactant times delta Gf of each reactant using the number of moles from the balanced chemical reaction Delta Gf for reaction = - RT lnK 242,090 J/mol = - (8.314 J/mol-K)(298.15 K) lnK K = Kp = 3.85E-43
How to calculate delta H, delta S, delta G and Kp for the following reaction?
d = delta dH(reaction) = sum of dH(formation) values for products - sum of dH(formation) values for reactants. dH(reaction) = [-235.1] - [-241.82 + 52.26] = -45.5 kJ dS(reaction) = sum of So values for products - sum of So values for reactants. dS(reaction) = [282.7 - (188.83 + 219.56) = -125.7 J/K dG = dH - T dS = -45.5 kJ - [298K x (- 0.1257 kJ/K)] = - 8.0 kJ = - 8000 J dGo = -RT ln(Kp) ln(Kp) = dGo / -RT ln(Kp) = -8,000 J / (- 8.315 J/K mol x 298 K) = 3.23 Kp = e^(3.23) = 25.2 Assume dSo and dHo don't change with temperature. (Use kelvin temperature in calculations: 250 C = 523 K) dGo = dHo - T dSo = -45.5 kJ - [523 K (-0.1257 kJ/K)] = +20.2 kJ/mol dGo = dHo - T dSo When dGo = 0, then T = dHo / dSo = -45.5 kJ / -0.1257 kJ/K = 362 K = 89 C I think I got everything. Good luck in your study of chemistry.
How much does the entropy of 1 kg of ice at 0 C change as it melt into water at 0 C, when it is strirred by a paddle wheel?
This is an example of irreversible adiabatic process. Here, the amount of heat transferred is zero (provided the ice is kept inside a thermally insulated container); but the entropy change is not zero. In other words, the equation[math]dS=\frac{\delta Q}{T}[/math] (1)is not valid in this case.[math]\Delta S[/math] can be calculated based on the fact that entropy is a state quantity. We can envisage a reversible process which has the same initial and final states as the process described in question and hence the same change in entropy - namely, melting ice by heat transfer. Here, we can easily substitute temperature and latent heat in equation 1.[math]\Delta S=\frac{\Delta Q}{T} = \frac{334 kJ}{273 K} = 1223 J/K[/math]
Why is the change in entropy of the vaporization of water 0 at 373 K?
The change in entropy of water cannot be 0 during vaporization. By definition, entropy is a measure of the "disorder" of a system or randomness.During vaporization the randomness of the system increases, so change in entropy will always have a positive value and cannot be 0.Change in Entropy at Vaporization is given by, [math]ΔHvap / T[/math]For example, when 1 g of liquid water evaporates at 373K(ΔHvap=40.63kJ/mol)[math]ΔSvap=ΔHvap / T[/math][math]ΔSvap[/math]=40.63×1000(J/mol) / 373 K[math]ΔSvap=[/math][math]109J/K−mol[/math]The above value is the change in entropy for 1 mol of water i.e. 18 grams of water. But we need the value for 1 gram of water.So, Entropy change for evaporation of 1 g of water =( 109×1) / 18=6.05 J/K
What is the free energy change, Delta G, in kilojoules for the reaction under these conditions?
In Part A, we saw that Delta G*=-242.1 for the hydrogenation of acetylene under standard conditions (298 K and all pressures equal to 1 ATM). In Part B, you will determine the Delta G for the reaction under a given set of nonstandard conditions. At 25 C the reaction from Part A has a composition as shown in the table below. Substance Pressure C2H2(g) 4.75 H2(g) 5.55 C2H6(g) 1.25×10−2 What is the free energy change, Delta G, in kilojoules for the reaction under these conditions?
Calculate the Gibbs Free Energy for the following reaction:?
Since temperature isn't mentioned, I am assuming that this is at 298 K (25 degree Celsius). The formula you want to use is delta G (products) - delta G (reactants) at standard temperature (298K). delta G for C2H6 (g) is -32.89 kJ/mol ' ' for O2 is 0 ' ' for CO2 (g) is -394.4 kJ/mol ' ' for H2O (g) is -273.13 kJ/mol Gibbs Free Energy (delta G) = ( (4 mol CO2 )* (-394.4 kJ/mol) + (6 mol H2O) * (-273.18 kJ/mol) ) - ( (2 mol C2H6) * (-32.89kJ/mol) + (7 mol O2) * (OkJ/mol) ) = -3150.6 kJ
Relationship between delta G and equilibrium constant K?
If we know the standard state free energy change, delta G, for a chemical process at some temperature T, we can calculate the equilibrium constant Keq for the process at that temperature using the relationship between delta Go and K. delta G= -RT ln Keq Rearrangement gives ln Keq= - delta G / RT Keq= e raised to the power (- delta G / RT) In this equation: number e = 2.71828 R = 8.314 J mol-1 K-1 or 0.008314 kJ mol-1 K-1. T is the temperature on the Kelvin scale.