Question

Use the ΔH°f and ΔH°rxn information provided to calculate ΔH°f for IF: ΔH°f (kJ/mol) IF7(g) + I2(g) → IF5(g) + 2 IF(g)

ΔH°rxn = -89 kJ

IF7(g) -941

IF5(g) -840

Answers:

(a) -190 KJ/mol

(b) 101 KJ/mol

(c) 24 KJ/mol

(d) -95 KJ/mol

(e) -146 KJ/mol

Answer #1

**IF _{7}(g) + I_{2}(g) → IF_{5}(g)
+ 2 IF(g) , ΔH°_{rxn} = - 89 kJ**

ΔH°rxn = (sum of ΔH°f products) - (sum of ΔH°f reactants)

given ΔH°rxn = -89 kJ

ΔH°f (IF7) = -941 kJ

ΔH°f(I_{2}) = 0 (elements in their standard state have
zero ΔH°f)

ΔH°f(IF5) = -840 kJ

ΔH°f(IF) = ?

so ΔH°rxn = [ ΔH°f(IF5) + 2 X ΔH°f(IF) ] - [ ΔH°f(IF7) +
ΔH°f(I_{2}) ]

-89 = [ -840 + 2 X ΔH°f(IF) ] - [ -941 + 0 ]

-89 = -840 + 2 X ΔH°f(IF) + 941

-89 = 101 + 2 X ΔH°f(IF)

2 X ΔH°f(IF) = -89 - 101 = -190

ΔH°f(IF) = -190/2 = -95

Answer is **(d) -95 kJ/ mol**

Use the ΔH°f and ΔH°rxn information provided to calculate ΔH°f
for IF: ΔH°f (kJ/mol) IF7(g) + I2(g) → IF5(g) + 2 IF(g) ΔH°rxn =
-89 kJ IF7(g) -941 IF5(g) -840

Use the following information to find ΔH°f
of gaseous HCl:
N2(g) + 3H2(g) →
2NH3(g)
ΔH°rxn = - 91.8 kJ
N2(g) + 4H2(g) +
Cl2(g) →
2NH4Cl(s)
ΔH°rxn = - 628.8 kJ
NH3(g) + HCl(g) →
NH4Cl(s)
ΔH°rxn = - 176.2 kJ

Use Hess's Law to calculate the enthalpy of reaction, ΔH rxn,
for the reaction in bold below given the following chemical steps
and their respective enthalpy changes. Show ALL work!
2 C(s) + H2(g) → C2H2(g) ΔH°rxn = ?
1. C2H2(g) + 5/2 O2(g) → 2CO2 (g) + H2O (l) ΔH°rxn = -1299.6
kJ
2. C(s) + O2(g) → CO2 (g) ΔH°rxn = -393.5 kJ
3. H2(g) + ½ O2(g) → H2O (l) ΔH°rxn = -285.8 kJ

Given the following data, calculate, ΔH rxn, ΔS rxn, and Δ rxn,
at 25° C for the equilibrium describe by the chemical equation.
What direction is the spontaneity of this system?
Mg (s) +HCl (aq) <---> H2 (g) +
MgCl2 (aq)
Mg(s)
HCl(aq)
H2 (g)
MgCl2(aq)
ΔH°f (Kj/mol)
0
-167.2
0
-801.3
S°(J/(mol K)
130.7
56.5
32.7
-24.0

Calculate the standard enthalpy change (ΔH⁰rxn ) for
the reaction of TiCl4(g) and
H2O(g) to form TiO2(s) and
HCl(g) given the standard enthalpies of formation
(ΔH⁰f ) shown in the table below. (Include the sign of the
value in your answer.)
kJ
Compound
ΔH⁰f
(kJ/mol)
TiCl4(g)
−763.2
H2O(g)
−241.8
TiO2(s)
−944.0
HCl(g)
−92.3

Calculate the ΔG∘rxn for the reaction using the following
information.
4HNO3(g)+5N2H4(l)→7N2(g)+12H2O(l)
ΔG∘f(HNO3(g)) = -73.5 kJ/mol;
ΔG∘f(N2H4(l)) = 149.3 kJ/mol;
ΔG∘f(N2(g)) = 0 kJ/mol;
ΔG∘f(H2O(l)) = -273.1 kJ/mol.
Calculate the for the reaction using the following
information.
= -73.5 ;
= 149.3 ;
= 0 ;
= -273.1 .
-312.9 kJ
+110.7 kJ
-954.7 kJ
-3.298 x 103 kJ
+2.845 x 103 kJ

Use standard enthalpies of formation to calculate ΔH∘rxn for the
following reaction: SO2(g)+12O2(g)→SO3(g) ΔH∘rxn =

1.Using the enthalpies of formation given below, calculate
ΔH°rxn in kJ, for the following reaction.
Report your answer to two decimal places in standard
notation.
H2S(g) + 2O2(g) → SO3(g) +
H2O(l)
H2S (g): -20.60 kJ/mol
O2 (g): 0.00 kJ/mol
SO3 (g): -395.77 kJ/mol
H2O (l): -285.83 kJ/mol
2. Calculate the amount of heat absorbed/released (in kJ) when
22.54 grams of SO3 are produced via the above
reaction.
Report your answer to two decimal places, and use appropriate signs
to...

Given the values of ΔH∘rxn, ΔS∘rxn, and
T below, determine ΔSuniv.
Part A
ΔH∘rxn= 117 kJ , ΔS∘rxn=− 263 J/K ,
T= 291 K .
ΔSuniv=
−1.3•102
J/K
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Incorrect; Try Again; 3 attempts remaining
Part B
ΔH∘rxn=− 117 kJ , ΔS∘rxn= 263 J/K ,
T= 291 K .
ΔSuniv=
J/K
SubmitMy AnswersGive
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Part C
ΔH∘rxn=− 117 kJ , ΔS∘rxn=− 263 J/K ,
T= 291 K.
ΔSuniv=
J/K
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Part D
ΔH∘rxn=− 117 kJ ,...

Given the values of ΔH∘rxn, ΔS∘rxn, and
T below, determine ΔSuniv.
Part A
ΔH∘rxn= 115 kJ , ΔS∘rxn=− 263 J/K ,
T= 301 K .
ΔSuniv=
J/K
SubmitMy AnswersGive
Up
Part B
ΔH∘rxn=− 115 kJ , ΔS∘rxn= 263 J/K ,
T= 301 K .
ΔSuniv=
J/K
SubmitMy AnswersGive
Up
Part C
ΔH∘rxn=− 115 kJ , ΔS∘rxn=− 263 J/K ,
T= 301 K .
ΔSuniv=
J/K
SubmitMy AnswersGive
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Part D
ΔH∘rxn=− 115 kJ , ΔS∘rxn=− 263 J/K ,
T= 557...

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