At equilibrium, \(1\) mole of \(CO\) is present
Hence, \(2-1=1\) moles of \(CO\) has reacted.
\(1\) mole of \(CO\) will react with \(1\) mole of \(Cl_2\) to from \(1\) mole of \(COCl_2\).
\(3-1=2\) moles of \(Cl_2\) remains at equilibrium
The equlibrium constant
\({K_c} = \frac{{[COC{l_2}]}}{{[CO][C{l_2}]}} = \frac{{\frac{{1\,mol}}{{5\,L}}}}{{\frac{{1\,mol}}{{5\,L}} \times \frac{{2\,mol}}{{5\,L}}}} = 2.5\)
$(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)}$
આ સંતુલન $\frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{(g)}$ માટે અચળાંક શું થશે?
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$(1)$ $ A_{2(g)} + 3B_{2(g)} $ $\rightleftharpoons$ $2AB_{3(g)} $ |
$(i)$ $(RT)^{-2}$ |
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$(2)$ $ A_{2(g)} + B_{2(g)} $ $\rightleftharpoons$ $ 2AB_{(g)}$ |
$(ii)$ $ (RT)^0$ |
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$(3)$ $A_{(s)} + 1.5 B_{2(g)} $ $\rightleftharpoons$ $ AB_{3(g)}$ |
$(iii)$ $(RT)^{1/2}$ |
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$(4)$ $AB_{2(g)} $ $\rightleftharpoons$ $AB_{(g)} + 0.5B_{(g)}$ |
$(iv)$ $(RT)^{-1/2}$ |
$(a) N_2O_4 $ $\rightleftharpoons$ $ 2NO_2$
$(b) 2SO_2 + O_2 $ $\rightleftharpoons$ $ 2SO_3$
$(c) X + Y $ $\rightleftharpoons$ $ 4Z$
$(d) A + 3B $ $\rightleftharpoons$ $ 7C$