At -17.4 °C the pressure equilibrium constant Kp 9.7x 10 for a certain reaction

Question

At -17.4 °C the pressure equilibrium constant Kp 9.7x 10 for a certain reaction Here are some facts about the reaction: . The net change in moles of gases is -1. The reaction is exothermic. . The constant pressure molar heat capacity Cn- 1.26 Jrmol -K OYes Using these facts, can you calculate Kp at -12. °C? ONo. If you said yes, then enter your answer at right. Round it to 2 significant digits. OYes, and Kp will be bigger If you said no, can you at least decide whether K, at O Yes, and K, will be 12. °C will be bigger or smaller than K at-17.4 °C? smaller O No.

Explanation / Answer

From Van;t Hoff equation

ln(K2/K1)= deltaH/R*(1/T1-1/T2) (1)

where K2 is equilibrium constant at T2 ( in K) and K1 is equilibrium constant at T1. DeltaH is enthalpy change of the reaction.

So in order to determine the Equilibrium constant at temperature T2=-12 deg.c, the data on enthalpy change of reaction is required. So the data is not adequate.

for exothermic reaction, deltaH is -ve, given T2=-12 deg.c= -12+273= 261K and T1=-17.4 deg.c= -17.4+273= 255.6, 1/T1-1/T2 is +ve.

deltaH is -ve.

So RHS is Eq.1 is -ve. this give less Kp values at -12 deg.c than at -17.4 deg.c. Kp will be smaller,

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