Class 11Chemistry1 Equilibrium: Chemical EquilibriumMix Examples- 6-1.Equilibrium (Chemical Equilibrium)
MediumFor the equilibrium:
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$; $K_{p} = 1.64 \ atm$ at $1000 \ K$.
$50 \ g$ of $CaCO_{3}$ in a $10 \ L$ closed vessel is heated to $1000 \ K$. The percentage of $CaCO_{3}$ that remains unreacted at equilibrium is:
(Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$; $K_{p} = 1.64 \ atm$ at $1000 \ K$.
$50 \ g$ of $CaCO_{3}$ in a $10 \ L$ closed vessel is heated to $1000 \ K$. The percentage of $CaCO_{3}$ that remains unreacted at equilibrium is:
(Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)
- A$50$
- B$20$
- C$40$
- D$60$
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Similar Questions
For a reaction $X + Y \rightleftharpoons 2 Z$,$1.0 \ mol$ of $X$,$1.5 \ mol$ of $Y$ and $0.5 \ mol$ of $Z$ were taken in a $1 \ L$ vessel and allowed to react. At equilibrium,the concentration of $Z$ was $1.0 \ mol \ L^{-1}$. The equilibrium constant of the reaction is $-\frac{x}{15}$. The value of $x$ is........
DifficultJEE MAIN 2020
View SolutionGiven the equilibrium constants for the following reactions:
$(1) \ CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)} \ ; \ K_1$
$(2) \ CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)} \ ; \ K_2$
What is the relationship between $K_1$ and $K_2$?
$(1) \ CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)} \ ; \ K_1$
$(2) \ CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)} \ ; \ K_2$
What is the relationship between $K_1$ and $K_2$?
Medium
View SolutionOne mole of $H_2O$ and one mole of $CO$ are taken in a $10 \ L$ vessel and heated to $725 \ K$. At equilibrium,$40 \%$ of water (by mass) reacts with $CO$ according to the equation:
$H_2O_{(g)} + CO_{(g)} \longleftrightarrow H_{2(g)} + CO_{2(g)}$
Calculate the equilibrium constant $(K_c)$ for the reaction.
$H_2O_{(g)} + CO_{(g)} \longleftrightarrow H_{2(g)} + CO_{2(g)}$
Calculate the equilibrium constant $(K_c)$ for the reaction.
Medium
View Solution$3.00 \ mol$ of $PCl_5$ kept in $1 \ L$ closed reaction vessel was allowed to attain equilibrium at $380 \ K$. If $1.59 \ mol$ of reactant was converted into the product at equilibrium,then $K_c$ is:
Medium
View SolutionFor the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,at equilibrium,the mole fraction of $PCl_5$ is $0.4$ and the mole fraction of $Cl_2$ is $0.3$. What will be the mole fraction of $PCl_3$?
Easy
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