Question

help please stressing

QUESTION 2 For the equilibrium 2PH3(e) = P2(8) +3H2(8), the equilibrium partial pressures are Ppzz-0.023 atm, Pp,-0.321 atm, and PH, -0.732 atm at 738K. Calculate Kp. *Please report 3 significant figures. Numbers only, No unit. No scientific notation. QUESTION 3 CH4(g) + 2 O2(g) = CO2(g) + 2 H2O(1) AH° -- 890 kJ Raising the temperature will result in shift the equilibrium to the right shift the equilibrium to the left O more information is needed Keq is increased O no effect in shifting the equilibrium

Answer 1

Question 2:

2PH3(g) = P2(g) + 3H2(g)

pP2 x (pH)3 (PH)2

P_PH3 = 0.023 atm

P_P2 = 0.321 atm

P_H2 = 0.732 atm

Therefore

0.321 x (0.7323 Ky = 2 (0.023)2

K, = 238atm?

Question asks for answer without units. So, Kp = 238

Question 3.

Raising the temperature shifts the equilibrium towards the endothermic side of the reaction.

In this case, since the forward reaction is exothermic (delta H is negative 890 kJ), therefore the reverse reaction will be endothermic (delta H will be positive 890 kJ)

Thus raising the temperature will shift the equilibrium towards the reverse reaction, i.e, towards the left.