(A) $\text{The dissociation reaction is: } 2HI(g) \rightleftharpoons H_2(g) + I_2(g)$$\text{Initial moles of } HI = 3.2 \text{ mol}$$\text{Degree of d

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(A) $\text{The dissociation reaction is: } 2HI(g) \rightleftharpoons H_2(g) + I_2(g)$$\text{Initial moles of } HI = 3.2 \text{ mol}$$\text{Degree of dissociation } (\alpha) = 22\% = 0.22$$\text{Moles of } HI \text{ dissociated} = \text{Initial moles} \times \alpha = 3.2 \times 0.22 = 0.704 \text{ mol}$$\text{Moles of } HI \text{ remaining at equilibrium} = \text{Initial moles} - \text{Dissociated moles}$$\text{Moles of } HI = 3.2 - 0.704 = 2.496 \text{ mol}$

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Degree of dissociation and Vapour density

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$3.2 \text{ moles of hydrogen iodide are heated in a closed vessel at } 444^{\circ}C \text{ until equilibrium is reached. If the degree of dissociation at this temperature is } 22\%, \text{ then the number of moles of hydrogen iodide remaining at equilibrium will be:}$

A
$2.496$
B
$1.87$
C
$2$
D
$4$

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6-1.Equilibrium (Chemical Equilibrium)

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Degree of dissociation and Vapour density

For a reaction at equilibrium $A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$,the relation between dissociation constant $(K)$,degree of dissociation $(\alpha)$ and equilibrium pressure $(p)$ is given by?

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The degree of dissociation of $PCl_5$ $(\alpha)$ obeying the equilibrium,$PCl_5 \rightleftharpoons PCl_3 + Cl_2$,is approximately related to the pressure at equilibrium by (given $\alpha << 1$):

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If the density of a gas at $NTP$ is $0.00445 \ g/mL$,then its vapor density and molecular weight are respectively:

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The degree of dissociation $(\alpha)$ of $PCl_5$ obeying the equilibrium $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is related to the pressure at equilibrium by

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If the vapor density of a volatile substance relative to $CH_4$ is $4$,then find the molecular mass of that substance.

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$3.2 \text{ moles of hydrogen iodide are heated in a closed vessel at } 444^{\circ}C \text{ until equilibrium is reached. If the degree of dissociation at this temperature is } 22\%, \text{ then the number of moles of hydrogen iodide remaining at equilibrium will be:}$

Similar Questions

For a reaction at equilibrium $A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$,the relation between dissociation constant $(K)$,degree of dissociation $(\alpha)$ and equilibrium pressure $(p)$ is given by?

The degree of dissociation of $PCl_5$ $(\alpha)$ obeying the equilibrium,$PCl_5 \rightleftharpoons PCl_3 + Cl_2$,is approximately related to the pressure at equilibrium by (given $\alpha << 1$):

If the density of a gas at $NTP$ is $0.00445 \ g/mL$,then its vapor density and molecular weight are respectively:

The degree of dissociation $(\alpha)$ of $PCl_5$ obeying the equilibrium $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is related to the pressure at equilibrium by

If the vapor density of a volatile substance relative to $CH_4$ is $4$,then find the molecular mass of that substance.

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