Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equation: X2(g) leftharpoons 2 X(g) The standard reaction Gibbs energy, Δr G°, 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, β text 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 Kp for this reaction at 298 K , in terms of β text equilibrium , is

Question

Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equation: X2(g) leftharpoons 2 X(g) The standard reaction Gibbs energy, Δr G°, 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, β text 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 Kp for this reaction at 298 K , in terms of β text equilibrium , is (A) (8 β text equilibrium 2/2-β text equilibrium ) (B) (8 β text equilibrium 2/4-β text equilibrium 2) (C) (4 β text equilibrium 2/2-β text equilibrium ) (D) (4 β text equilibrium 2/4-β text equilibrium 2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/b854da4dee989f0cc6583742caab2905-.png" /> Kp=((pX)2/(pX2)) ; pX=((βe q/1-(βe q)2+βe q)) P text total =((βe q/1+(βe q)2)) P text total pX2=((1-(βe q/2)/1+(βe q)2)) P text total Kp=([((βe q/1+(βe q)2)) P text total ]2/((1-(βe q)2/1+(βe q)2)) P text total )=((βe q2/1-(βe q2)4)) P text total =((4 βe q2/4-βe q2)) P text total =((4 βe q2/4-βe q2)) × 2=(8 βe q2/4-βe q2)

$\frac{8 β _{equilibrium}^{2}}{2 - β _{equilibrium}}$

$\frac{8 β _{equilibrium}^{2}}{4 - β _{equilibrium}^{2}}$

$\frac{4 β _{equilibrium}^{2}}{2 - β _{equilibrium}}$

$\frac{4 β _{equilibrium}^{2}}{4 - β _{equilibrium}^{2}}$

2−βequilibrium ​8βequilibrium 2​​

4−βequilibrium 2​8βequilibrium 2​​

2−βequilibrium ​4βequilibrium 2​​

4−βequilibrium 2​4βequilibrium 2​​

$\frac{8 β _{equilibrium}^{2}}{2 - β _{equilibrium}}$

$\frac{8 β _{equilibrium}^{2}}{4 - β _{equilibrium}^{2}}$

$\frac{4 β _{equilibrium}^{2}}{2 - β _{equilibrium}}$

$\frac{4 β _{equilibrium}^{2}}{4 - β _{equilibrium}^{2}}$