At $1024 \mathrm{~K}$, the equilibrium constant $K_c$ for the reaction $\mathrm{H}_2(g)+\mathrm{Br}_2(g) \rightleftharpoons 2 \mathrm{HBr}(g)$ is $1.6 \times 10^{\wedge} 5$. If 10.0 bar of $\mathrm{HBr}$ is introduced into a sealed container at $1024 \mathrm{~K}$, what will be the equilibrium pressures of $\mathrm{H}_2, \mathrm{Br}_2$, and $\mathrm{HBr}$ ?

Question

At $1024 \mathrm{~K}$, the equilibrium constant $K_c$ for the reaction $\mathrm{H}_2(g)+\mathrm{Br}_2(g) \rightleftharpoons 2 \mathrm{HBr}(g)$ is $1.6 \times 10^{\wedge} 5$. If 10.0 bar of $\mathrm{HBr}$ is introduced into a sealed container at $1024 \mathrm{~K}$, what will be the equilibrium pressures of $\mathrm{H}_2, \mathrm{Br}_2$, and $\mathrm{HBr}$ ?, $4.0 \times 10^5$ bar, $5.0 \times 10^5$ bar, $1.0 \times 10^5$ bar, $1.0 \times 10^5$ bar, $2.5 \times 10^5$ bar, $2.5 \times 10^5$ bar, $3.3 \times 10^5$ bar, $3.3 \times 10^5$ bar, $3.3 \times 10^5$ bar, $5.0 \times 10^5$ bar, $4.0 \times 10^5$ bar, $1.0 \times 10^5$ bar

MARKS App · Accepted Answer

Correct Option is : (B) $1.0 \times 10^5$ bar, $2.5 \times 10^5$ bar, $2.5 \times 10^5$ bar

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