1.

Thermal decomposition of gaseous `X_(2)`to gaseous X at 298 K takes place accoeding to the following equation: `X_(2_(g)hArr2X(g)` is positive. At the start of the reaction, there is one mole of `X_(2)` and no X. As the reaction proceeds, the number of moles of X formed is given by `beta.` The `beta_("equulibrium")` is the number of moles of X formed at equlibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given: The standard reaction Gibbs energy, `Delta_(r)G^(@),` of this reaction `R=0.08L barK^(-1)mol^(-1)`) The equlibrium constatn `K_(p)` for this reaction at 298 K, in terms of `beta_("equlibrium,")` isA. `(8beta_("equlibrium")^(2))/(2-beta_("equlibrium"))`B. `(8beta_("equlibrium")^(2))/(4-beta_("equlibrium")^(2))`C. `(4beta_("equlibrium")^(2))/(2-beta_("equlibrium"))`D. `(4beta_("equlibrium")^(2))/(4-beta_("equlibrium")^(2))`

Answer» Correct Answer - B
`underset(1)(X_(2))(g)hArr2XC(g)`
`-(beta_(e))/(2)" "beta_(e)`
Total number of moles at equlibrium
`implies1-(beta_(e))/(2)+beta_(e)implies1+(beta_(e))/(2)`
`K_(P)=((P_(x))^(2))/(P_(x_(2)))`
`=((((beta_(e)xx2)/(1+(beta_(e))/(2)))^(2))/((1-(beta_(e))/(2))xx2))/(1+(beta_(e))/(2))=(2beta_(e)^(2))/(1-(beta_(e^(2)))/(4))impliesK_(p)=(8beta_(e)^(2))/(4-beta_(e)^(2))`


Discussion

No Comment Found

Related InterviewSolutions