Home
Class 12
CHEMISTRY
What fraction of the surface of a crysta...

What fraction of the surface of a crystal of `Cd` at `T = 298 K` consists of vacancies? Assume that the energy needed to form a vacancy `= 0.5 Delta_("sub")H^(Θ)`. For `Cd(s), Delta_("sub")H^(Θ) = 112.0 kJ mol^(-1)`.

A

`6.02 xx 10^(-22)`

B

`6.02xx 10^(22)`

C

`1.5 xx 10^(-10)`

D

`1.50xx 10^(10)`

Text Solution

Verified by Experts

Fraction is expressed in terms of
`(n)/(N) = exp (-( E)/(RT)) = 1.52 xx10^(-10)`
Given E= `112.01xx10^(2) "Jmol"^(-1) xx(50)/(100) `
R `= 8.313 "Jmol"^(-1) K^(-1)`
T = 298 K
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • THE SOLID STATE

    VMC MODULES ENGLISH|Exercise LEVEL-2 (Numerical Value Type for JEE Main )|15 Videos
  • THE SOLID STATE

    VMC MODULES ENGLISH|Exercise JEE Main (Archive )|39 Videos
  • THE SOLID STATE

    VMC MODULES ENGLISH|Exercise LEVEL-1|75 Videos
  • SURFACE CHEMISTRY

    VMC MODULES ENGLISH|Exercise PRACTICE EXERCISE|9 Videos
  • THEORY OF SOLUTIONS

    VMC MODULES ENGLISH|Exercise JEE Advanced (Archive)|31 Videos

Similar Questions

Explore conceptually related problems

Calculate the entropy change of n -hexane when 1mol of it evaporates at 341.7K (Delta_(vap)H^(Theta) = 290.0 kJ mol^(-1))

Calculate the entropy change in surroundings when 1.00 mol of H_(2)O(l) is formed under standard conditions, Delta_(r )H^(Θ) = -286 kJ mol^(-1) .

Standard molar enthalpy of formation, Delta_(f) H^(Θ) is just a special case of enthalpy of reaction, Delta_(r) H^(Θ) . Is the Delta_(r) H^(Θ) same as standard molar enthalpy of formation ? Given reason for your answer. CaO (s) + CO_(2) (g) rarr CaCO_(3) (s) , Delta_(f) H^(Θ) = - 178.3 kJ mol^(-1)

The combustion of 1 mol of benzene takes place at 298 K and 1 atm . After combustion, CO_(2)(g) and H_(2)O(l) are produced and 3267.0 kJ of heat is librated. Calculate the standard entalpy of formation, Delta_(f)H^(Θ) of benzene Given: Delta_(f)H^(Θ)CO_(2)(g) = -393.5 kJ mol^(-1) Delta_(f)H^(Θ)H_(2)O(l) = -285.83 kJ mol^(-1) .

Calculate the lattice energy from the following data (given 1 eV = 23.0 kcal mol^(-1) ) i. Delta_(f) H^(ɵ) (KI) = -78.0 kcal mol^(-1) ii. IE_(1) of K = 4.0 eV iii. Delta_("diss")H^(ɵ)(I_(2)) = 28.0 kcal mol^(-1) iv. Delta_("sub")H^(ɵ)(K) = 20.0 kcal mol^(-1) v. EA of I = -70.0 kcal mol^(-1) vi. Delta_("sub")H^(ɵ) of I_(2) = 14.0 kcal mol^(-1)

Use the following data to calculate Delta_("lattice") H^(Θ) " for " NaBr. Delta_("Sub")H^(Θ) for sodium metal = 108.4 kJ mol^(-1) , ionisation enthalpy of sodium = 496 kJ mol^(-1) , electron gain enthalpy of bromine = -325 kJ mol^(-1) , bond dissociation enthalpy of bromine = 192 kJ mol^(-1), Delta_(f) H^(Θ) " for " NaBr(s) = - 360.1 kJ mol^(-1)

Calculate the enthalpy change for the process C Cl_(4)(g) rarr C(g)+4Cl(g) and calculate bond enthalpy of C-Cl in C Cl_(4)(g) . Delta_(vap)H^(Θ)(C Cl_(4))=30.5 kJ mol^(-1) Delta_(f)H^(Θ)(C Cl_(4))=-135.5 kJ mol^(-1) Delta_(a)H^(Θ)(C )=715.0 kJ mol^(-1) , where Delta_(a)H^(Θ) is enthalpy of atomisation Delta_(a)H^(Θ)(Cl_(2))=242 kJ mol^(-1)

Calculate the equilibrium constant for the following reaction at 298 K and 1 atm pressure.C(graphite)+H2O(l)=CO(g)+H2(g) Given : 'Delta_f H^@ , [H_2O (l)] = -286.0 kJ mol^(-1) , Delta_f H^@ [ CO (g)] = - 110.5 kJ mol^(-1) , DeltaS^@ at 298 K for the reaction =252.6 J k^(-1) mol^(-1) . Gas constant R = 8.31 Jk^(-1) mol^(-1) .

Calculate the free energy change when one mole of sodium chloride is dissolved in water at 298 K. (Given : Lattice energy of NaCl = - 777.8 kJ mol^(-1) , Hydration energy of NaCl = 774.1 kJ mol^(-1) and DeltaS " at" 298 K = 0.043 kJ mol^(-1) .

Will the reaction, I_(2)(s) +H_(2)S(g) rarr 2HI(g) +S(s) proceed spontaneously in the forward direction of 298K Delta_(f)G^(Theta)HI(g) = 1.8 kJ mol^(-1), Delta_(f)G^(Theta)H_(2)S(g) = 33.8 kJ mol^(-1) ?