(D) The reaction is: $Fe^{3+} + SCN^{-} \rightleftharpoons FeSCN^{2+}$Initial moles: $[Fe^{3+}] = 3.1 \ M$,$[SCN^{-}] = 3.2 \ M$,$[FeSCN^{2+}] = 0 \ M

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(D) The reaction is: $Fe^{3+} + SCN^{-} \rightleftharpoons FeSCN^{2+}$Initial moles: $[Fe^{3+}] = 3.1 \ M$,$[SCN^{-}] = 3.2 \ M$,$[FeSCN^{2+}] = 0 \ M$At equilibrium,$[FeSCN^{2+}] = 3.0 \ M$Change in moles: $[Fe^{3+}] = 3.1 - 3.0 = 0.1 \ M$,$[SCN^{-}] = 3.2 - 3.0 = 0.2 \ M$Equilibrium constant $K_c = \frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^{-}]}$$K_c = \frac{3.0}{(0.1)(0.2)} = \frac{3.0}{0.02} = 150$

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Law of equilibrium and Equilibrium constant

Easy

$3.1 \ mol$ of $FeCl_3$ and $3.2 \ mol$ of $NH_4SCN$ are added to $1 \ L$ of water. At equilibrium,$3.0 \ mol$ of $FeSCN^{2+}$ is formed. The equilibrium constant $K_c$ for the reaction is:
$Fe^{3+} + SCN^{-} \rightleftharpoons FeSCN^{2+}$

A
$6.66 \times 10^{-3}$
B
$0.3$
C
$3.3$
D
$150$

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6-1.Equilibrium (Chemical Equilibrium)

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Law of equilibrium and Equilibrium constant

What is $K_{c}$ for the following equilibrium when the equilibrium concentration of each substance is: $[SO_{2}] = 0.60 \, M, [O_{2}] = 0.82 \, M$ and $[SO_{3}] = 1.90 \, M?$
$2SO_{2(g)} + O_{2(g)} \longleftrightarrow 2SO_{3(g)}$

Easy

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$4.5$ moles each of hydrogen and iodine are heated in a sealed $10 \ L$ vessel. At equilibrium,$3$ moles of $HI$ are found. The equilibrium constant for ${H_2}_{(g)} + {I_2}_{(g)} \rightleftharpoons 2HI_{(g)}$ is

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$HI$ was heated in a closed tube at $440\,^{\circ}C$ until equilibrium was obtained. At this temperature,$22\%$ of $HI$ was dissociated. The equilibrium constant for this dissociation will be:

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For the reaction $2 SO_2 + O_2 \rightleftharpoons 2 SO_3$,the units of $K_c$ are

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If an equilibrium process equation is multiplied by a factor $n$,how does the equilibrium constant change? Explain with an example.

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$3.1 \ mol$ of $FeCl_3$ and $3.2 \ mol$ of $NH_4SCN$ are added to $1 \ L$ of water. At equilibrium,$3.0 \ mol$ of $FeSCN^{2+}$ is formed. The equilibrium constant $K_c$ for the reaction is:$Fe^{3+} + SCN^{-} \rightleftharpoons FeSCN^{2+}$

Similar Questions

What is $K_{c}$ for the following equilibrium when the equilibrium concentration of each substance is: $[SO_{2}] = 0.60 \, M, [O_{2}] = 0.82 \, M$ and $[SO_{3}] = 1.90 \, M?$$2SO_{2(g)} + O_{2(g)} \longleftrightarrow 2SO_{3(g)}$

$4.5$ moles each of hydrogen and iodine are heated in a sealed $10 \ L$ vessel. At equilibrium,$3$ moles of $HI$ are found. The equilibrium constant for ${H_2}_{(g)} + {I_2}_{(g)} \rightleftharpoons 2HI_{(g)}$ is

$HI$ was heated in a closed tube at $440\,^{\circ}C$ until equilibrium was obtained. At this temperature,$22\%$ of $HI$ was dissociated. The equilibrium constant for this dissociation will be:

For the reaction $2 SO_2 + O_2 \rightleftharpoons 2 SO_3$,the units of $K_c$ are

If an equilibrium process equation is multiplied by a factor $n$,how does the equilibrium constant change? Explain with an example.

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$3.1 \ mol$ of $FeCl_3$ and $3.2 \ mol$ of $NH_4SCN$ are added to $1 \ L$ of water. At equilibrium,$3.0 \ mol$ of $FeSCN^{2+}$ is formed. The equilibrium constant $K_c$ for the reaction is:
$Fe^{3+} + SCN^{-} \rightleftharpoons FeSCN^{2+}$

What is $K_{c}$ for the following equilibrium when the equilibrium concentration of each substance is: $[SO_{2}] = 0.60 \, M, [O_{2}] = 0.82 \, M$ and $[SO_{3}] = 1.90 \, M?$
$2SO_{2(g)} + O_{2(g)} \longleftrightarrow 2SO_{3(g)}$