$5 \ \text{moles}$ of $SO_2$ and $5 \ \text{moles}$ of $O_2$ are allowed to react to form $SO_3$ in a closed vessel. At the equilibrium stage,$60\%$ of $SO_2$ is used up. The total number of moles of $SO_2$,$O_2$,and $SO_3$ in the vessel now is:

  • A
    $10$
  • B
    $8.5$
  • C
    $10.5$
  • D
    $3.9$

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$(i)$ Find $K_p$.
$(ii)$ At $0.1 \ bar$ pressure and $310 \ K$,what is the percentage of $N_2O_4$ decomposed?

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In a closed vessel at $448^{\circ} C$,$0.5 \ mol$ of $H_2$ and $0.5 \ mol$ of $I_2$ react to form hydrogen iodide.
Reaction: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,$K_c = 50$.
$(i)$ Calculate the moles of $I_2$ that remain unreacted at equilibrium.
$(ii)$ Calculate $K_p$.

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For the reaction $X_{(g)} + Y_{(g)} \rightleftharpoons Z_{(g)}$ at $550 \ K$,the value of $K_c$ is $10^{-4} \ mol^{-1} \ L$. If at equilibrium $[X] = \frac{1}{2}[Y] = \frac{1}{2}[Z]$,then the value of $[Z]$ at equilibrium will be:

The equilibrium constants for the following three reactions $(i)$,$(ii)$,and $(iii)$ are given as:
$(i)$ $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)} \quad K_1$
$(ii)$ $CH_{4(g)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + 3H_{2(g)} \quad K_2$
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