Discussion Forum

Thermal decomposition of gaseous X₂, to gaseous X at 298 K takes place according to the following equation

$X_{2}(g)\rightleftharpoons2X(g)$

The standard reaction Gibbs energy, Δ_rG⁰, of this reaction is positive. At the start of the reaction, there is one mole of X₂ and no X. As the reaction proceeds, the number of moles of X formed is given by β. Thus, βequilibrium is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given, R = 0.083 L bar K^-1 mol^-1)

The incorrect  statement among the following reaction , is

Answer:

Explanation:

a)          $K_{p}= \frac{4px^{2}}{(4-x^{2})}=px^{2}(\because 4>>>x)$

$\therefore x\alpha \sqrt{\frac{1}{p}}$

If p decreases, x increases. Equilibrium is shifted in the forward  side. Thus, statement(a) is correct

(b)       At the start of the reaction , Q=0 where , Q is the reaction quotient

   $\triangle G= \triangle G^{0}+2.303RT \log Q$

   since   $\triangle G^{0}>0$  , thus  $\triangle G$ is -ve

Hence, dissociation take place spontaneously,

 Thus (b) is correct

 (c) If we use x= 0.7 and p=2 bar,

  then  $K_{p}=\frac{4\times 2(0.7)^{2}}{[4-(0.7)^{2}]}=1.16>1$

Thus, (c) is incorrect

(d) At equilibium, $\triangle G=0$

$\therefore$      $\triangle G ^{0}=-2.303RT\log K_{p}$

Since , $\triangle G ^{0}=+ve$

$Hence, K_{p}< 1,K_{C}= \frac{K_{p}}{(RT)}$

Then $K_{C}< 1$ , Thus (d) is correct