InterviewSolution
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InterviewSolution
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`K_(p) " for the reaction " N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) " at "400^(@)C " is " 1*64 xx10^(-4) atm^(-2)." Find " K_(c). " Also calculate " DeltaG^(@) " using " K_(p) and K_(c) " values and interpret the difference ".` |
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Answer» `N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g)` `Delta n= 2-4 = -2` `K_(p) = K_(c) (RT)^(Delta n)` `1*64 xx 10^(-4) " atm"^(-2) = K_(c) (0*0821) L " atm "K^(-1) mol^(-1) xx 673 K ) ^ (-2) " "(T=400 + 273 K = 673 K)` or `K_(c) = (1*64 xx 10^(-4) "atm"^(-2))/((0*821 xx 673 " atm"^(-1) mol^(-1))^(2))=0*5372 " mol"^(2)L^(-2)` Now, `Delta G^(@) = - 2*303" RT " log K` If `K=K_(p)` ` Delta G^(@) = - 2*303 xx (8*314 " JK"^(-1) mol^(-1)) (673 K) xx log (1*64 xx 10^(-4))` `=-2*303 xx 8*314 xx 673 xx (-3* 7852) " J "mol^(-1) = +48*78 " kj "mol^(-1)` If `K=K_(c)` , `Delta G^(@)= - 2*303 xx (8*314 xx 673 xx (-0*27) " J "mol^(-1) = + 3479 " J "mol^(-1)` Interpreting the difference .`DeltaG^(@)` is the free energy change when all the reactants and products are in their standard state. In case of `K_(p)`, standard state pressure is used which is 1 atm (or now 1 bar) wheras in case of `K_(c)` , standard concentration is used which is 1 mol `L^(-1)` (as already explained on page 7/22). |
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