Use the values of Δ H f ° in Appendix 4 to calculate ∆ H º for the following reactions. a. b. Ca 3 ( PO 4 ) ( s ) + 3 H 2 SO 4 ( l ) → 3 CaSO 4 ( s ) + 2 H 3 PO 4 ( l ) c. NH 3 ( g ) + HCl ( g ) → NH 4 Cl ( s )
Use the values of Δ H f ° in Appendix 4 to calculate ∆ H º for the following reactions. a. b. Ca 3 ( PO 4 ) ( s ) + 3 H 2 SO 4 ( l ) → 3 CaSO 4 ( s ) + 2 H 3 PO 4 ( l ) c. NH 3 ( g ) + HCl ( g ) → NH 4 Cl ( s )
Use the values of
Δ
H
f
°
in Appendix 4 to calculate ∆Hº for the following reactions.
a.
b.
Ca
3
(
PO
4
)
(
s
)
+
3
H
2
SO
4
(
l
)
→
3
CaSO
4
(
s
)
+
2
H
3
PO
4
(
l
)
c.
NH
3
(
g
)
+
HCl
(
g
)
→
NH
4
Cl
(
s
)
a)
Expert Solution
Interpretation Introduction
Interpretation: Standard enthalpy change has calculated for given reaction.
Concept introduction
Standard Enthalpy change (ΔH0): The heat change when molar quantities of reactants as specified by chemical equation to form a product at standard conditions. Standard condition: 250C and 1 atmosphere pressure.
Interpretation: Standard enthalpy change has calculated for given reaction.
Concept introduction
Standard Enthalpy change (ΔH0): The heat change when molar quantities of reactants as specified by chemical equation to form a product at standard conditions. Standard condition: 250C and 1 atmosphere pressure.
Standard state for given compound in the reaction are,
Substance and state ΔHf0kJ/ mole
NH3(g) -46
O2(g) 0
CH4(g) -75
HCN(g) -135.1
H2O(g) -242
Ca(PO4)2(s) -4126H2SO4(l) -814
CaSO4(s) -1433
H3PO4(s) -1267
HCl(g) -92
NH4Cl(s) -314
The standard state of ammonia gas, oxygen, methane, hydrogen cyanide and water vapour are given. By substituting the values in the standard enthalpy change equation the standard enthalpy change for the reaction calculated as -940kJ/mol.
The standard state of some compounds which present in the reaction are given. By substituting the values in the standard enthalpy change equation, the standard enthalpy change for the reaction calculated as -265kJ.
c)
Expert Solution
Interpretation Introduction
Interpretation: Standard enthalpy change has calculated for given reaction.
Concept introduction
Standard Enthalpy change (ΔH0): The heat change when molar quantities of reactants as specified by chemical equation to form a product at standard conditions. Standard condition: 250C and 1 atmosphere pressure.
Answer to Problem 79E
NH3(g)+HCl(g)→NH4Cl(s)ΔH0=-176kJ
Explanation of Solution
Given data
Standard state for given compound in the reaction are,
Substance and state ΔHf0kJ/ mole
NH3(g) -46
O2(g) 0
CH4(g) -75
HCN(g) -135.1
H2O(g) -242
Ca(PO4)2(s) -4126H2SO4(l) -814
CaSO4(s) -1433
H3PO4(s) -1267
HCl(g) -92
NH4Cl(s) -314
To calculate standard enthalpy change.
The balanced equation is,
NH3(g)+HCl(g)→NH4Cl(s)ΔH0=-176kJ
ΔH0= ∑np ΔHf0,product−∑nr ΔHf0,reactant
=[1(-314kJ)]-[1(-46kJ)+1(-92kJ)]
=-176kJ
The standard state of some compounds which present in the reaction are given. By substituting the values in the standard enthalpy change equation, the standard enthalpy change for the reaction calculated as -176kJ.
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For a titration of 40.00 mL of 0.0500 M oxalic acid H2C2O4 with 0.1000 M KOH, calculate the pH at each of the following volume of KOH used in the titration: 1) before the titration begin;2) 15 mL; 3) 20 mL; 4) 25 mL; 5) 40 mL; 6) 50 mL. Ka1 = 5.90×10^-2, Ka2 = 6.50×10^-5 for oxalic acid.
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