The Born Haber cycle below represents the energy changes occurring at 298K when KH is formed from its elementsv - - H atomisation K - 90 kJ-molw - - H ionisation K - 418 kJ-molx - - H dissociation H - 436 kJ-moly - - H electron affinity H - 78 kJ-molz - - H lattice KH - 710 kJ-molCalculate the value of - H showing all your working-

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Question

The Born Haber cycle below represents the energy changes occurring at 298K when $K H$ is formed from its elements
v : ${Δ H}_{a t o m i s a t i o n}$ $K = 90 k J / m o l$
w : ${Δ H}_{i o n i s a t i o n}$ $K = 418 k J / m o l$
x : ${Δ H}_{d i s s o c i a t i o n}$ $H = 436 k J / m o l$
y : ${Δ H}_{e l e c t r o n a f f i n i t y}$ $H = 78 k J / m o l$
z : ${Δ H}_{l a t t i c e}$ $K H = 710 k J / m o l$
Calculate the value of $Δ$ $H$ showing all your working.

124 kJ/mol

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-124 kJ/mol

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124 J/mol

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None of these

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Solution

The correct option is A -124 kJ/mol
Born-Haber Cycle
It is a series of steps (chemical processes) used to calculate the lattice energy of ionic solids, which is difficult to determine experimentally. You can think of BH cycle as a special case of Hess's law which states that the overall energy change in a chemical process can be calculated by breaking down the process into several steps and adding the energy change from each step.
${Δ H}_{r}$ =2 $\times$ 90 + 2 $\times$ 418 + 436 - 2 $\times$ 78 - 2 $\times$ 710
${Δ H}_{r}$ = $- 124 k J / m o l e$

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Similar questions

Q. The Born Haber cycle below represents the energy changes occurring at 298K when

K H

is formed from its elements
v :

{Δ H}_{a t o m i s a t i o n}

K = 90 k J / m o l

w :

{Δ H}_{i o n i s a t i o n}

K = 418 k J / m o l

x :

{Δ H}_{d i s s o c i a t i o n}

H = 436 k J / m o l

y :

{Δ H}_{e l e c t r o n a f f i n i t y}

H = 78 k J / m o l

z :

{Δ H}_{l a t t i c e}

K H = 710 k J / m o l

Out of v to y the most endothermic is___.

The Born Haber cycle below represents the energy changes occurring at 298K when

K H

is formed from its elements
v :

{Δ H}_{a t o m i s a t i o n}

K = 90 k J / m o l

w :

{Δ H}_{i o n i s a t i o n}

K = 418 k J / m o l

x :

{Δ H}_{d i s s o c i a t i o n}

H = 436 k J / m o l

y :

{Δ H}_{e l e c t r o n a f f i n i t y}

H = 78 k J / m o l

z :

{Δ H}_{l a t t i c e}

K H = 710 k J / m o l

On complete reaction with water,

0.1 g

K H

gave a solution requiring 25

{c m}^{3}

of 0.1M

H C l

for neutralisation.Calculate the relative atomic mass of potassium from this information.

The Born Haber cycle below represents the energy changes occurring at 298K when KH is formed from its elements :
v :

{Δ H}_{a t o m i s a t i o n}

K = 90 kJ/mol
w :

{Δ H}_{i o n i s a t i o n}

K = 418 kJ/mol
x :

{Δ H}_{d i s s o c i a t i o n}

H = 436 kJ/mol
y :

{Δ H}_{e l e c t r o n a f f i n i t y}

H = 78 kJ/mol
z :

{Δ H}_{l a t t i c e}

KH = 710 kJ/mol

In terms of the letters v to z the expression for
(i)

Δ

H for the reaction

2 K

(s) +

H_{2}

(g)

⟶

2 K H

(s).
will be

Δ

$H =_____?

Q. The Born Haber cycle below represents the energy changes occurring at 298K when KH is formed from its elements
v :

{Δ H}_{a t o m i s a t i o n}

K = 90 kJ/mol
w :

{Δ H}_{i o n i s a t i o n}

K = 418 kJ/mol
x :

{Δ H}_{d i s s o c i a t i o n}

H = 436 kJ/mol
y :

{Δ H}_{e l e c t r o n a f f i n i t y}

H = 78 kJ/mol
z :

{Δ H}_{l a t t i c e}

KH = 710 kJ/mol

Out of v to y the most exothermic is___.

Q. The Born Haber cycle below represents the energy changes occurring at 298K when KH is formed from its elements
v :

{Δ H}_{a t o m i s a t i o n}

K = 90 kJ/mol
w :

{Δ H}_{i o n i s a t i o n}

K = 418 kJ/mol
x :

{Δ H}_{d i s s o c i a t i o n}

H = 436 kJ/mol
y :

{Δ H}_{e l e c t r o n a f f i n i t y}

H = 78 kJ/mol
z :

{Δ H}_{l a t t i c e}

KH = 710 kJ/mol

In terms of the letters v to z the expression for

{Δ H}_{e l e c t r o n a f f i n i t y}

of H is

{Δ H}_{e l e c t r o n a f f i n i t y}

is __.
I

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