For the reaction `CH_(4(g))+2O_(2(g))hArrCO_(2(g))+2H_(2)O_(l)`:
(`DeltaH=-170.8 kJmol^(-1)`). Which of the following statement is not true?

Options

A. At equilibrium the concentrations of `CO_2(g) and H_2O(I)` are not equal.
B. The equilibrium constant for the reaction is given by `K_p=([CO_2])/([CH_4][O_2])`
C. Addition of `CH_4(g) or O_2(g)` at equilibrium will cause a shift to the right
D. The reaction is exothermic

Answer

B

Explanation

For the reaction, `CH_4(g) 2O_2(g)hArrCO_2(g)+2H_2O(I)`
`DeltaH_r)=-170.8kJmol^(-)`
This equilibrium is an example of haterogenous chemical equilibrium. Hence for it.
`([CO_2])/([CH_4][O_2]^2)` ....(ii)
(equilibrium constant on the basis of concentration) and
`K_p=(P_(co_2))/(p_(ch._4)xxp_(o_2^2))` (ii)
(equilibrium constant according to partial pressure)
Thus, in this contentration of `CO_2(g) and H_2O(I)` are not equal at equilibrium
The equilibrium constant `K_(p)=([CO_2])/([CH_4][O_2])` is not correct expression.
On adding `CH_4` (g) or `O_2` (g) at equilibrium `K_c` will be decreased according to expression (i) but `K_c` remains constant at constant temperature for a reaction. Hence, on addition of `CH_4` or` O_2` equilibrium will cause to the right.
Combustan reaction is an example of exothermic reaction.