Answer:
-1,103.39KJ/mol
Explanation:
We use the subtract the standard enthalphies of formation of the reactants from that of the products. It must be taken into consideration that the enthalpy of formation of elements and their molecules alone are not taken into consideration. Hence, what we would be considering are the standard enthalpies of formation of H2S, H2O and SO2.
In places where we have more than one mole, we multiply by the number of moles as seen in the balanced chemical equations.
The standard enthalpies of the molecules above are as follows:
H2S = -20.63KJ/mol
H2O = -285.8KJ/mol
SO2 = -296.84KJ/mol
O2 = 0KJ/mol
ΔrH⦵ = [2ΔfH⦵(H2O) + 2 ΔfH⦵(SO2)] − [ΔfH⦵(H2S) + 3
ΔfH⦵(O2)]
ΔrH⦵ =[(2 × -285.8) + (2 × -296.84)]
-[ 3 × -20.63)]
= (-571.6 - 593.68 + 61.89) = -1,103.39KJ/mol
The standard Gibbs free energy of formation of ZnO from Zn is lower than that of CO2 from CO. Therefore, CO cannot reduce ZnO to Zn. Hence, Zn is not extracted from ZnO through reduction using CO
Changing the volume increases the area that the molecules collide with so the force is spread over a larger area.
When writing an ionic compound formula, a "molecular" form is used. The formula is made with allowance for ion charges.
For example,
Ca²⁺ and NO₃⁻ ⇒ Ca(NO₃)₂
Al³⁺ and SO₄²⁻ ⇒ Al₂(SO₄)₃
Answer:
ΔH = -470.4kJ
Explanation:
It is possible to sum 2 or more reactions to obtain the ΔH of the reaction you want to study (Hess's law). Using the reactions:
1. CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)ΔH = −414kJ
2. 6C2H2(g) + 3CO2(g) + 4H2O(g) → 5CH2CHCO2H(g)ΔH = 132kJ
6 times the reaction 1.
6CaC2(s) + 12H2O(l) → 6C2H2(g) + 6Ca(OH)2(s)ΔH = −414kJ*6 = -2484kJ
This reaction + 2:
6CaC2(s) + 3CO2(g) + 16H2O(l) → + 6Ca(OH)2(s) + 5CH2CHCO2H(g) ΔH = -2484kJ + 132kJ = -2352kJ
As we want to calculate the net change enthalpy in the formation of just 1 mole of acrylic acid we need to divide this last reaction in 5:
6/5CaC2(s) + 3/5CO2(g) + 16/5H2O(l) → + 6/5Ca(OH)2(s) + CH2CHCO2H(g) ΔH = -2352kJ / 5
<h3>ΔH = -470.4kJ</h3>
