Question

4. For which of the two reactions below will the equilibrium constant increase with increasing tempat 250z(8) + 0.8 - 25O3(8) AH = - 47 kcal/mol 2H (9) H:() + 1:(8) AH = + 3.21 kcal/mol 5. Before travelling from Orlando to Miami, a student accidentally inflates his car tires with a known concentration of HX gas instead of air. Some of the HX immediately dissociates to form H, and X2 as shown in the reaction below. Based on what you know about frictional heat and about chemical equilibrium, will the concentration of HX continue to decrease as the trip progress, or will it remain the same? Explain. Assume that there are no leaks or reactions between HX and the tires. 2HX (g) H(g) + X:(8) AH = +0.51 kcal/mol 6. How would the addition of Cl to the following reaction aflect the position of equilibrium? Ag(NH3)2 Ag + 2NH, not change since Cl" is a spectator ion that is not included in the (a) The position of the equilibrium will not change since Cl' is a spectator ion that is reaction on with the positively charged Ag(NH3)2. This will effectively (b) The CI will form a complex ion with the positively char remove Ag(NH3)2 from solution and force the reaction to the lo sitively charged Ag. This will effectively remove Ag! (c) The Cl' will form a precipitate with the positively char from solution and shift the equilibrium to the right

Answer 1

5)

According to Le Chatelier's Principle, the equilibrium constant for an endothermic reaction ($\Delta H>0$) increases with an increase in temperature.

Explanation

The variation of equilibrium constant K_eq with temperature T is given by the Van't Hoff equation,

$\frac{d\ln{K_{eq}}}{dT}=\frac{\Delta H}{RT^2}$

Here, R is the universal gas constant. Thus, if $\Delta H>0$ then the Right hand side is positive i.e.

$\frac{d\ln{K_{eq}}}{dT}=(+)$

This means that increasing the T will increase ln (K_eq) and hence K_eq.

Therefore, the answer will be equation 2 for which $\Delta H>0$ i.e.

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$