Question

For the reaction $CH_{4}(g)+2O_2(g) {<=>}CO_2(g)+2H_2O(\ell) \Delta H_{r} =-170.8\,KJ\,mol^{-1}$ Which of the following statements is not true:

• A. At equilibrium, the concentrations of $C O_2(g)$ and $H_2O (l)$ are not equal
• B. The equilibrium constant for the reaction is given by $ K_p = \frac{ [ CO_2 ] }{ [ CH_4 ] [ O_2 ] } $
• C. Addition of $CH_4$ (g)or $O_2$ (q)at equilibrium will cause a shift to the right
• D. The reaction is exothermic

Answer 1

The Correct Option is B

For the reaction,

$ CH_4 \, ( g) + 2 O_2 ( g) \leftrightharpoons CO_2 \, ( g) + 2 H_2 0 (I), $
$ \triangle_r H = - 170.8 k J mol^{ - 1 } $

This equilibrium is an example of heterogeneous chemical equilibrium. Hence, for it

$ K_c = \frac{ [CO_2 ] }{ [ CH_4 ] [ O_2 ]^2 } $ $20\, mm\,\,\,\,\dots(i)$

(equilibrium constant on the basis of concentration)

and $ K_p = \frac{ \rho_{ co_2} }{ \rho_{ CH_4 } \times \rho_{0_2^2} } $ $20\,mm\,\,\,\,\dots(ii)$

(equilibrium constant according to partial pressure)

Thus, in this concentration of $ CO_2 \, ( g ) $ and $ H_2 O ( I) $ are not equal at equilibrium.

The equilibrium constant $ ( K_p) = \frac{ [ CO_2 ] }{ [CH_4 ] [ O_2 ] } $ is not correct expression.

On adding $CH_4$ (g) or $0_2$ (g) at equilibrium, $K_c$ will be decreased according to expression (i) but $K_c$ remains constant at constant temperature for a reaction, so for maintaining the constant value of $K_c$ , the concentration of C$O_2$ will increased in same order. Hence, on addition of C$H_4 $ or $ O_2$ equilibrium

will cause to the right.

Combustion reaction is an example of exothermic reaction