△f G° for (NH4Cl,s) at 310 K. Given △fH° (NH4Cl,s) = -314.5 kJ/mol;
△tCp=O is -
S°N2(g) = 192 JK-1mol-1
S°H2(g) = 130.5 JK-1mol-1
S°Cl2(9) = 233 JK-1mol-1
S°NH4Cl(S) = 99.5 JK-1mol-1
(All given data at 300 K)
1. -168.56 kJ/mol
2. -426.7 kJ/mol
3. -202.3 kJ/mol
4. None of the above
Using listed informations, calculate △fG° (in kJ/mol) at 27°C Co3O4(s)+4CO(g)→3Co(s)+4CO2(g) Given: At 300 K
△fH° (kJ/mol) -891, -110.5, 0.0,-393.5 ;
S0 (J/K-mol) 102.5, 197.7, 30.0, 213.7
1. -214.8
2. -195.0
3. -200.3
4. -256.45
Determine △U° at 300 K for the following reaction using the listed enthalpies of
reactions
4CO(g)+8H2(g)→3CH4(g)+CO2(g)+2H2O(l)
C(graphite)+12O2(g)→CO(g); △H01=-110.5 kJ
CO(g)+12O2(g)→CO2(g); △H02=-282.9 kJ
H2(g)+12O2(g)→H2O(l); △H03=-285.8 kJ
C(graphite)+2H2(g)→CH4(g) △H04=-74.8 kJ
1. -653.5 kJ
2. -686.2 kJ
3. -747.4kJ
4. None of these
△fH° (in kJ/mol) for Cr2O3 from △f G° and the S° values provided at 27°C for the given reaction is -
4Cr(s)+3O2(g) →2Cr2O3(s);△r G° = -2093.4 kJ/mol S° (J/k mol) : S (Cr,s) = 24; S0 (O2,g) = 205; S° (Cr2O3,s) = 81
1. -2258.1 kJ/mol
2. -1129.05 kJ/mol
3. -964.35 kJ/mol
4. None of the above
The heat produced (in kJ) when 224 gm of CaO is completely converted to CaCO3 by reaction with CO2 at 27°C in a container of fixed volume is -
Given:△fH0 (CaCO3,s) = -1207 kJ/mol; △fH0(CaO,s) = -635 kJ/mol △fH0(CO2,g) = - 394 kJ/mol; [Use R = 8.3 K-1 mol-1]
1. 302.04 kJ
2. 721.96 kJ
3. 712 J
4. 71 kJ
Enthalpy of neutralization of H3PO3 acid is - 106.68 kJ/mol using NaOH. If enthalpy of neutralization of HCl by NaOH is -55.84 kJ/mol. ΔIonization of H3PO3 will be
1. 50.84 kJ/mol
2. 5 kJ/mol
3. 2.5 kJ/mol
4. None of the above
The enthalpy of neutralization of weak monoprotic acid (HA) in 1 M solution with a
strong base is -55.95 kJ/mol. If the unionized acid is required 1.4 kJ/mol heat for it’s complete ionization and enthalpy of neutralization of the strong monobasic acid
with a strong monobasic base is -57.3 kJ/mol. What is the % ionization of the weak acid in molar solution
1. 1%
2. 3.57%
3. 35.7%
4. 10%
Bond enthalpy of C-C and C-H bond enthalpy (in kJ/mol) is -
Given:
△fH0(C2H6,g) = -85 kJ/mol, △fH0(C3H8,g) = - 104 kJ/mol; △subH0(C,s) = 718 kJ/mol, B.E. (H-H) = 436 kJ/mol
1. 414, 345
2. 345,414
3. 287,404.5
4. None of the above.
Consider the following data : △fH°((N2H4,l) = 50 kJ/mol, △f H° (NH3,g) = -46 kJ/mol B.E. (N-H) = 393 kJ/mol and B.E. (H-H) = 436 kJ/mol △vapH(N2H4,l) = 18 kJ/mol The N-N bond energy in N2H4 is-
1. 236 kJ/mol
2. 154 kJ/mol
3. 190 kJ/mol
4. None of the above
From given following equations and △H0 values, determine the enthalpy of reaction at 298 K for the reaction:
C2H4(g) + 6F2(g)→2F4(g)+4HF(g)
H2(g)+F2(g)→2HF(g); △H01 = -537 kJ
C(s)+2F2(g)→CF4(g); △H02 = -680 kJ
2C(s)+2H2(g)→C2H4(g); △H03 = 52 kJ
1. -1165
2. -2486
3. +1165
4. +2486
Consider the following reactions:
C(s) + (g) + x kJ
CO(g) + ----> + y kJ
The heat of formation of CO(g) is-
1. -(x+y) kJ/mol
2. (x-y) kJ/mol
3. (y-x) kJ/mol
4. None of the above
The fat, C57H1O6(s) is metabolized via the following reaction. Given the enthalpies of formation, calculate the energy (kJ) liberated when 1.0 g of this fat reacts
1. -37.98
2. -40.4
3. -33.4
4. -30.2
Determine the enthalpy of formation for , using the following enthalpies of reaction :
kJ/mol
kJ/mol
1. | -383 kJ/mol | 2. | -187 kJ/mol |
3. | -49 kJ/mol | 4. | None of the above |
Enthaply of neutralization of HCl by NaOH is 55.84 kJ/mol and by is -51.34 kJ/mol. The enthalpy of ionization of is :
1. 10.18 kJ/mol
2. 4.5 kJ/mol
3. -4.5 kJ/mol
4. None of the above