Question

at - 13.5°c the pressure equilibrium constant ( k_p = 4.4 ) for a certain reaction.
here are some facts about the reaction:

• if the reaction is run at constant pressure, 104. kj/mol of heat are absorbed.
• the net change in moles of gases is - 1.
• some of the reactants are liquids and solids.

using these facts, can you calculate ( k_p ) at - 2.8°c? yes. no.
if you said yes, then enter your answer at right. round it to 2 significant digits.
if you said no, can you at least decide whether ( k_p ) at - 2.8°c will be bigger or smaller than ( k_p ) at - 13.5°c? yes, and ( k_p ) will be bigger. yes, and ( k_p ) will be smaller. no.

Explanation:

Step1: Determine the reaction type

Since heat is absorbed at constant pressure, the reaction is endothermic ($\Delta H>0$).

Step2: Analyze the effect of temperature change on $K_p$ for endothermic reactions

For an endothermic reaction, when the temperature increases ($T_2=-2.8^{\circ}C = 270.35\ K$, $T_1=-13.5^{\circ}C=259.65\ K$, $T_2 > T_1$), according to the van't Hoff equation $\ln\frac{K_{p2}}{K_{p1}}=\frac{\Delta H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$. Since $\Delta H>0$, $T_2 > T_1$, then $\frac{1}{T_1}-\frac{1}{T_2}>0$, so $\ln\frac{K_{p2}}{K_{p1}}>0$, which means $K_{p2}>K_{p1}$.

Answer:

Yes, and $K_p$ will be bigger.