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-molOn complete reaction with water- 0-1 g of KH gave a solution requiring 25 cm 3 of 0-1M HCl for neutralisation-Calculate the relative atomic mass of potassium from this information-

The correct option is A 39 Meq. of KH = Meq. of HCl 0.1 / E _ KH × 1000 = 25 × 0.1 Valency factor (05955) of K is 1 hence E _ K ...

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Standard XII

Physics

Nuclear Fission

The Born Habe...

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 : ${Δ H}_{l a t t i c e}$ $K H = 710 k J / m o l$
On complete reaction with water, $0.1 g$ of $K H$ 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.

39

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40

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41

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None of these

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Solution

The correct option is A $39$
Meq. of KH = Meq. of $H C l$
$\frac{0.1}{E_{K H}} \times 1000 = 25 \times 0.1$
Valency factor (05955) of $K$ is 1 hence
$E_{K}$ = $M_{K}$ $M_{K}$ =39
$E_{K H}$ =40 $E_{K H}$ = $E_{K}$ =
40= $E_{K}$ +1 $E_{K H}$
$E_{K}$ $\Rightarrow$ 39

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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

Calculate the value of

Δ

H

showing all your working.

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 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

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 =_____?

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The correct option is A 39Meq. of KH = Meq. of HCl0.1EKH×1000=25×0.1Valency factor (05955) of K is 1 henceEK=MK MK=39EKH=40 EKH=EK=40=EK+1 EKHEK ⇒ 39

The correct option is A $39$
Meq. of KH = Meq. of $H C l$
$\frac{0.1}{E_{K H}} \times 1000 = 25 \times 0.1$
Valency factor (05955) of $K$ is 1 hence
$E_{K}$ = $M_{K}$ $M_{K}$ =39
$E_{K H}$ =40 $E_{K H}$ = $E_{K}$ =
40= $E_{K}$ +1 $E_{K H}$
$E_{K}$ $\Rightarrow$ 39