(A) The value of $\Delta n$ is calculated as $\Delta n = \sum n_p - \sum n_r$,where $n_p$ is the number of moles of gaseous products and $n_r$ is the

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(A) The value of $\Delta n$ is calculated as $\Delta n = \sum n_p - \sum n_r$,where $n_p$ is the number of moles of gaseous products and $n_r$ is the number of moles of gaseous reactants. For reaction $(1): C_2H_{6(g)} \rightleftharpoons C_2H_{4(g)} + H_{2(g)}$,$\Delta n = (1 + 1) - 1 = +1$. For reaction $(2): N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$,$\Delta n = 2 - (1 + 1) = 0$. For reaction $(3): H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,$\Delta n = 2 - (1 + 1) = 0$. Therefore,$\Delta n$ is positive for reaction $(1)$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Kp and Kc Relationship

Medium

For which of the following reactions will the value of $\Delta n = \sum n_p - \sum n_r$ be positive? $(1)$ $C_2H_{6(g)} \rightleftharpoons C_2H_{4(g)} + H_{2(g)}$ $(2)$ $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ $(3)$ $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$

A
For $(1)$
B
For $(2)$
C
For $(3)$
D
None of these

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

$3.00 \, mol$ of $PCl_5$ kept in $1 \, L$ closed vessel was allowed to attain equilibrium at $380 \, K$. If $K_C = 1.80$,calculate the equilibrium concentrations of $PCl_3$ and $Cl_2$ respectively.

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One mole of $PCl_{5(g)}$ was heated in a $1 \ L$ closed flask at $500 \ K$. At equilibrium,$0.1 \ mole$ of $Cl_{2(g)}$ was formed. What is its $K_{p}$ (in $atm$)? (Given $R=0.082 \ L \ atm \ mol^{-1} \ K^{-1}$)

MediumAP EAMCET 2023

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$1 \ mol$ $N_2$ and $3 \ mol$ $H_2$ are heated at $473 \ K$ and $100 \ atm$ pressure. At equilibrium,the number of moles of $NH_3$ is $0.5 \ mol$. Calculate the equilibrium constant $K_p$ of the given reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

Medium

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At equilibrium,the concentrations are $[N_2] = 3.0 \times 10^{-3} \ M$,$[O_2] = 4.2 \times 10^{-3} \ M$,and $[NO] = 2.8 \times 10^{-3} \ M$ in a sealed vessel at $800 \ K$ and $1 \ atm$ pressure. What will be $K_p$ for the given reaction?
$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$

DifficultAIIMS 2018

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For the reaction $SO_2 + \frac{1}{2} O_2 \rightleftharpoons SO_3$,if we write $K_p = K_c(RT)^x$,then $x$ becomes

EasyWBJEE 2009

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For which of the following reactions will the value of $\Delta n = \sum n_p - \sum n_r$ be positive? $(1)$ $C_2H_{6(g)} \rightleftharpoons C_2H_{4(g)} + H_{2(g)}$ $(2)$ $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ $(3)$ $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$

Similar Questions

$3.00 \, mol$ of $PCl_5$ kept in $1 \, L$ closed vessel was allowed to attain equilibrium at $380 \, K$. If $K_C = 1.80$,calculate the equilibrium concentrations of $PCl_3$ and $Cl_2$ respectively.

One mole of $PCl_{5(g)}$ was heated in a $1 \ L$ closed flask at $500 \ K$. At equilibrium,$0.1 \ mole$ of $Cl_{2(g)}$ was formed. What is its $K_{p}$ (in $atm$)? (Given $R=0.082 \ L \ atm \ mol^{-1} \ K^{-1}$)

$1 \ mol$ $N_2$ and $3 \ mol$ $H_2$ are heated at $473 \ K$ and $100 \ atm$ pressure. At equilibrium,the number of moles of $NH_3$ is $0.5 \ mol$. Calculate the equilibrium constant $K_p$ of the given reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

At equilibrium,the concentrations are $[N_2] = 3.0 \times 10^{-3} \ M$,$[O_2] = 4.2 \times 10^{-3} \ M$,and $[NO] = 2.8 \times 10^{-3} \ M$ in a sealed vessel at $800 \ K$ and $1 \ atm$ pressure. What will be $K_p$ for the given reaction?$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$

For the reaction $SO_2 + \frac{1}{2} O_2 \rightleftharpoons SO_3$,if we write $K_p = K_c(RT)^x$,then $x$ becomes

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At equilibrium,the concentrations are $[N_2] = 3.0 \times 10^{-3} \ M$,$[O_2] = 4.2 \times 10^{-3} \ M$,and $[NO] = 2.8 \times 10^{-3} \ M$ in a sealed vessel at $800 \ K$ and $1 \ atm$ pressure. What will be $K_p$ for the given reaction?
$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$