The Born Haber cycle below represents the energy changes occurring at 298K when KH is formed from its elements -v - - 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-molIn terms of the letters v to z the expression fori -H for the reaction 2Ks - H 2g - 2KHs-will be - H -

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Hess' Law

The Born Habe...

Question

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 =_____?

(v + w + x + y + z)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(2v + 2w + x + 2y + 2z)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(v + w + 2x + y +2 z)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(v + 2w + x + y + z)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B (2v + 2w + x + 2y + 2z)
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 $Δ$ $H = (2 v + 2 w + x + 2 y + 2 z)$ .

Suggest Corrections

Similar questions

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

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

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___.

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

Calculate the value of

Δ

H

showing all your working.

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.

Join BYJU'S Learning Program

The correct option is B (2v + 2w + x + 2y + 2z)In terms of the letters v to z the expression for(i) ΔH for the reaction2K(s) + H2(g) ⟶ 2KH(s).will be ΔH=(2v+2w+x+2y+2z).

The correct option is B (2v + 2w + x + 2y + 2z)
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 $Δ$ $H = (2 v + 2 w + x + 2 y + 2 z)$ .