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Byju's Answer

Standard XII

Chemistry

Degree of Dissociation

For the gaseo...

Question

For the gaseous reaction :
$C_{2} H_{2} + D_{2} O ⇌ C_{2} D_{2} + H_{2} O$ ,

$Δ H$ is $530 c a l$ . At $25 K_{p} = 0.82$ . If X moles of $C_{2} D_{2}$ will be formed, if 1 mole $C_{2} H_{2}$ and 2 mole of $D_{2} O$ are put together at a total pressure of 1 atm at $100$ , then $100 X$ is:

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Solution

The correct option is B 54
The relationship between the equilibrium constants at two different temperatures is as given below.
$l o g \frac{k_{2}}{k_{1}} = \frac{Δ H}{2.303 R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$
Substitute values in the above expression.
$l o g \frac{k_{2}}{0.82} = \frac{530}{2.303 \times 2} (\frac{1}{298} - \frac{1}{373})$
$l o g \frac{k_{2}}{0.82} = 0.0776$
$\frac{k_{2}}{0.82} = 1.1957$
$k_{2} = 0.98$
The initial and the equilibrium number of moles of various species are listed below.

$C_{2} H_{2}$
$D_{2} O$ $C_{2} D_{2}$
Initial moles
1
2
0
equilibrium moles
$1 - X$ $2 - X$ $X$
$\frac{x (3 - X)}{(1 - x) (2 - x)} = 0.98$
$X^{2} - 6.0594 X + 3.0198 = 0$
$x = 0.54$
$100 x = 54$

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	$C_{2} H_{2}$	$D_{2} O$	$C_{2} D_{2}$
Initial moles	1	2	0
equilibrium moles	$1 - X$	$2 - X$	$X$

For the gaseous reaction :C2H2+D2O⇌C2D2+H2O, ΔH is 530 cal. At 25Kp=0.82. If X moles of C2D2 will be formed, if 1 mole C2H2 and 2 mole of D2O are put together at a total pressure of 1 atm at 100, then 100X is:

For the gaseous reaction :
$C_{2} H_{2} + D_{2} O ⇌ C_{2} D_{2} + H_{2} O$ ,

$Δ H$ is $530 c a l$ . At $25 K_{p} = 0.82$ . If X moles of $C_{2} D_{2}$ will be formed, if 1 mole $C_{2} H_{2}$ and 2 mole of $D_{2} O$ are put together at a total pressure of 1 atm at $100$ , then $100 X$ is: