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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?
`Delta_("sub") H^(@) (Cd) = 113.02 kJ mol^(-1)` Energy needed in form a vacancy is appoximately 60% of `Delta _("sub") H^(0)`

A

`1.50 xx 10^(16)`

B

`6.02 xx 10^(-23)`

C

`6.02 xx 10^(22)`

D

`1.30 xx 10^(-12)`

Text Solution

Verified by Experts

The correct Answer is:
d
Fraction is expressed in terms of
`(n)/(N) = exp (-(E )/(KT))`
`E = 113.02 xx 10^(3) 1 mA^(-1) = (60)/(100)`
`R = 8.3143 JmA^(-1) K^(-1)`
`T = 298 K`
`(n)/(N) = ((113.02 xx 10^(3) xx 0.6)/(8.3143 xx 298))`
`= exp (- 27.369) - 1.30 + 10^(-12)`
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Knowledge Check

  • The activation energy for the forward reaction X rarr Y is 60 kJ mol^(-1) and Delta H is -20 kJ mol^(-1) . The activation energy for the reverse reaction is

    A
    `40 kJ mol^(-1)`
    B
    `60 kJ mol^(-1)`
    C
    `80 kJ mol^(-1)`
    D
    `20 kJ mol^(-1)`

  • The change of energy on freezing 1.00 kg of liquid water of 0^(@)C and 1 atm is : (Delta H _("ice"))_("fusion") = 6. 01 kJ // mol

    A
    `236.7 kJ kg ^(-1)`
    B
    `236.4 kJ kg ^(-1)`
    C
    `-333.4 kJ kg ^(-1)`
    D
    `- 236.7 kJ kg ^(-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) ltbvrgt v. EA of I = -70.0 kcal mol^(-1) vi. Delta_("sub")H^(ɵ) of I_(2) = 14.0 kcal mol^(-1)

    A
    `+14.1 kcal mol^(-1)`
    B
    `-14.1 kcal mol^(-1)`
    C
    `-141 kcal mol^(-1)`
    D
    `+141 kcal mol^(-1)`
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