HARD
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Which of the following parameter cannot be estimated by using Born - Haber cycle?

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Important Questions on Thermodynamics

HARD

An athlete is given 100 g of glucose C6H12O6 for energy. This is equivalent to 1800 kJ of energy. The 50% of this energy gained is utilized by the athlete for sports activities at the event. In order to avoid storage of energy, the weight of extra water he would need to perspire is _____g (Nearest integer) Assume that there is no other way of consuming stored energy.

Given : The enthalpy of evaporation of water is 45 kJ mol-1

Molar mass of C,H&O are $12.1$ and 16 g mol-1.

EASY
At 298 K, the enthalpy of fusion of a solid X is 2.8 kJ mol-1 and the enthalpy of vaporisation of the liquid X is 98.2 kJ mol-1. The enthalpy of sublimation of the substance X in kJ mol-1 is ______. (in nearest integer)
MEDIUM

The change in enthalpy ΔH in kJmol-1 for the reaction is Mg+2 FMgF2

Given: EA of F=328 kJ mol-1,IE1 of Mg=737 kJ mol-1,IE2 of Mg=1451 kJ mol-1

EASY
If Cs+O2gCO2g,ΔH=-X,
COg+12O2gCO2g,ΔH=-Y,
Calculate ΔfH for COg formation
HARD
Enthalpy of fusion and enthalpy of vaporization for water respectively are 6.01 kJ mol-1 and 45.07 kJ mol-1 at 0°C what is enthalpy of sublimation at 0°C?
EASY
Given

NOg+O3gNO2g+O2g     H=-198.9  kJ/mol

O3g3/2O2g         H=-142.3 kJ/mol

O2g2Og          H=+495.0 kJ/mol

The enthalpy change (H) for the following reaction is

NOg+OgNO2g
HARD

The Born-Haber cycle for KCl is evaluated with the following data:
ΔfHΘ for KCl=436.7kJmol1;ΔsubHΘ for K=89.2kJmol1;

Δionization HΘ for K=419.0kJ mol 1;Δelectron gain HΘ for Cl(g)=348.6kJmol1

Δbond HΘ for Cl2=243.0kJmol1
The magnitude of lattice enthalpy of KCl in kJmol-1 is (Nearest integer)

MEDIUM
The enthalpy change for the conversion of 12Cl2 (g) to Cl-(aq) is (-)
kJmol-1 (Nearest integer)
Given : Δdis HCl2( g)0=240kJmol-1.
ΔegHCl(g)o=-350kJmol-1,
ΔhydHCl(g)-o=-380kJmol-1
MEDIUM
The elements with the highest and the lowest enthalpy of atomisation respectively, for first row transition elements are:
HARD

For one mole of NaCl(s) the lattice enthalpy is

Na(s)+12Cl2g+108.4 kJ/molNa(s)+12Cl2g+495.6 kJ/molNa(g)++12Cl2g121 kJ/mol

Na(g)++Clg-348.6 kJ/molNa(g)++Cl(g)-ΔH0 altice Nacl (solid) -411.2 kJ/molNa(s)+12Cl2g

EASY
Given that, the molar combustion enthalpy of benzene, cyclohexane and hydrogen are x, y and z, respectively, the molar enthalpy of hydrogenation of benzene to cyclohexane is
MEDIUM
Enthalpy of sublimation of iodine is 24 cal g-1 at  200°C . If specific heat of I2 (s) and I2 (vap) are 0.055 and 0.031 cal g-1K-1 respectively, then enthalpy of sublimation of iodine at 250°C in cal g-1 is:
MEDIUM
The enthalpy of combustion of propane, graphite and dihydrogen at 298 K are: -2220.0 kJ mol-1, -393.5 kJ mol-1 and -285.8 kJ mol-1, respectively. The magnitude of enthalpy of formation of propane C3H8 is _____ kJmol-1. (Nearest integer)
MEDIUM
Given:

i Cgraphite+O2gCO2g; ΔrHΘ=x kJ mol-1

ii Cgraphite+12O2gCOg ; ΔrHΘ=y kJ mol-1

iii COg+12O2gCO2g; ΔrHΘ=z kJ mol-1

Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct?
MEDIUM

At 25°C, the enthalpy of the following processes are given:

H2( g)+O2( g)2OH(g)ΔHo=78 kJ mol-1

H2( g)+1/2O2( g)H2O(g)ΔH0=-242 kJ mol-1

H2( g)2H(g)ΔHo=436 kJ mol-1

1/2O2( g)O(g)ΔH0=249 kJ mol-1

What would be the value of X for the following reaction? (Nearest integer)

H2OgHg+OHg Ho=X kJmol-1

MEDIUM

Consider the following data

Heat of combustion of H2(g)=241.8 kJ mol1

Heat of combustion of C(s)=393.5 kJ mol1

Heat of combustion of  C2H5OH(l)=1234.7 kJ mol1

The heat of formation of C2H5OH(l) is () _____ kJ mol1 (Nearest integer).

MEDIUM

Given

(A) 2COg+O2g2CO2g ΔH1o=-x kJ mol-1

(B) Cgraphite+O2gCO2g ΔH2o=-y kJ mol-1

The Ho for the reaction Cgraphite+12O2gCOg is

MEDIUM
The ionization enthalpy of Na+ formation from Nag is 495.8 kJ mol-1, while the electron gain enthalpy of Br is -325.0 kJ mol-1. Given the lattice enthalpy of NaBr is -728.4 kJ mol -1 . The energy for the formation of NaBr ionic solid is -_____10-1 kJ mol-1
MEDIUM
Lattice enthalpy and enthalpy of solution of NaCl are 788kJmol-1and  4kJmol -1, respectively. The hydration enthalpy of NaCl is:
 
EASY
The enthalpy change on freezing of 1 mol of water at 5°C to ice at -5°C is:

(Given Δfus H=6 kJ mol-1 at 0°C,

CpH2O, l=75.3 J mol-1K-1

CpH2O, s=36.8 J mol-1K-1 )