Thermal decomposition of gaseous X_(2) to gaseous X at 298 K takes place according to the following equation : X_(2)(g) 2X(g) The standard reaction Gibbs energy, Delta_(r)G°, of this reaction is positive. At the start of the reaction, there is one moe of X_(2) and no X. As the reaction proceeds, the number of moles of X formed is given by beta. Thus, beta_("equilibrium") 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 equilibrium constant K_(p) for this reaction at 298 K, in terms of beta_("equilibrium"), is

Thermal decomposition of gaseous X_(2) to gaseous X at 298 K takes pla

1.	Thermal decomposition of gaseous X_(2) to gaseous X at 298 K takes place according to the following equation : X_(2)(g) 2X(g) The standard reaction Gibbs energy, Delta_(r)G°, of this reaction is positive. At the start of the reaction, there is one moe of X_(2) and no X. As the reaction proceeds, the number of moles of X formed is given by beta. Thus, beta_("equilibrium") 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 equilibrium constant K_(p) for this reaction at 298 K, in terms of beta_("equilibrium"), is
Answer» `(8beta_("EQUILIBRIUM")^(2))/(2-beta_("equilibrium"))` `(8beta_("equilibrium")^(2))/(4-beta_("equilibrium")^(2))` `(4beta_("equilibrium")^(2))/(2-beta_("equilibrium"))` `(4beta_("equilibrium")^(2))/(4-beta_("equilibrium")^(2))` ANSWER :B