4. Calculate the standard heat of formation of calcium carbide (CaC) from its elements in kJ/mol using thefollowing thermochemical equations.(hint come up with the overall equation first)Ca(s) + 2H₂O ⇒ Ca(OH)2(s) + H₂(g)2C(s) + O2(g) → 2COCaO + H₂O⇒ Ca(OH)2()2H2(g) + O2(g) → 2H₂O)CaO) + 3C,,→ CaCs) + COΔΗ°= - 414.79 kJΔΗ°= - 221.00 kJΔΗ°= - 65.19 kJΔΗ°= - 571.80 kJAH°= + 462.30 kJ1 Selected Standard Molar Enthalpies of FormationAH₁AH;AH₁Substance (kJ/mol) Substance (kJ/mol) Substance (kJ/mol)Al₂O3(s)-1675.7 HBr(g)CaCO3(s)-1207,6 HCl(g)CaCl₂(s)Ca(OH)₂(s)CCL₂(E)CCL4(8)CHCI ()CH₂(g)C₂H₂(g)C₂H₁(g)C₂H₂(g)CzHs(g)CH()CH₂OH(E)C₂H₂OH()CO(g)CO₂(g)COCI₂(g)+49.1 Fe₂O₂(8)-239.2 Fe,O4(s)-277.6 FeCl₂(s)CH3COOH(E) –484.3 FeCls(s)-110.5 FeS (8)CS₂(e)CS₂(g)CrCla(g)Cu(NO₂)₂ (8)-795.4 HF(g)-985.2 HCN(g).-128.2 H₂O(E)-95.7 H₂O(g)CuO(s)CuCl(s)CuCl₂(s)-134.1 H₂O₂(e)-74.6 HNO (0)+227.4 H₂PO4(s)+52.4 H₂S(g)-84.0 H₂SO4(e)-103.8 FeO(s)–36.3|NH(8)-92.3 N₂H4(e)-273.3 NH Cl(s)+135.1 NH₂NO(s)-285.8 NO(g)-393.5 PbCl₂(s)-219.1 MgCl₂(s)+89.0 MgO(s)+116.7 Mg(OH)2(s)-556.5 HgS(s)-241.8 NO₂(g)-187.8 N₂O(g)-174.1 N₂O(g)1284.4 PH,(g)-20.6 PCl₂(g)814.0 P,O,(s)-272.0 P4O₁0(s)-824.2 KBr(s)-1118.4 KCl(s)-341.8 KCIO,(s)-399.5 KOH(s)-178.2 AgaCO₂ (s)-359.4 AgCl(s)-641.3 AgNO,(s)-601.6 Ag S(s)-924.5 SFB(g)-58.2 SO₂(g)-302.9 NaCl(s)-411.2 SO₂(g)-157.3 NaOH(s)-425.6 SnCl₂(s)-137.2 Na₂CO3(s)-1130.7 SnCl4(e)-220.1-45.9+50.6-314.4-365.6+91.3+33.2+81.6+11.1+5.4-287.0-2144.3-2984.0-393.8-436.5-397.7-424.6-505.8-127.0-124.4-32.6-1220.5-296.8-395.7-325.1-511.3Note: The enthalpy of formation of an element in its standard state isdefined as zero.

4. Recall the given reactions, Ca s + 2 H2O l → CaOH2 s + H2 g ∆H° = -…

4. Calculate the standard heat of formation of calcium carbide (CaC) from its elements in kJ/mol using the following thermochemical equations. (hint come up with the overall equation first) Ca+ 2H₂O ⇒ Ca(OH)2 + H₂ 2C+0⇒ 2CO CaO + H₂O⇒ Ca(OH)200 2H₂ + O⇒ 2H₂O CaO + 3C = CaC + CO ΔΗ°= - 414.79 kJ ΔΗ°= - 221.00 kJ ΔΗ°= - 65.19 kJ ΔΗ°= - 571.80 kJ ΔΗ°= + 462.30 kJ

4. Calculate the standard heat of formation of calcium carbide (CaC) from its elements in kJ/mol using the following thermochemical equations. (hint come up with the overall equation first) Ca+ 2H₂O ⇒ Ca(OH)2 + H₂ 2C+0⇒ 2CO CaO + H₂O⇒ Ca(OH)200 2H₂ + O⇒ 2H₂O CaO + 3C = CaC + CO ΔΗ°= - 414.79 kJ ΔΗ°= - 221.00 kJ ΔΗ°= - 65.19 kJ ΔΗ°= - 571.80 kJ ΔΗ°= + 462.30 kJ

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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