Answer:
(a) The equilibrium partial pressure of BrCl (g) will be greater than 2.00 atm.
Explanation:
Q is the coefficient of the reaction and is calculated the same of the way of the equilibrium constant, but using the concentrations or partial pressures in any moment of the reaction, so, for the reaction given:
Q = (pBrCl)²/(pBr₂*pCl₂)
Q = 2²/(1x1)
Q = 4
As Q < Kp, the reaction didn't reach the equilibrium, and the value must increase. As we can notice by the equation, Q is directly proportional to the partial pressure of BrCl, so it must increase, and be greater than 2.00 atm in the equilibrium.
The partial pressures of Br₂ and Cl₂ must decrease, so they will be smaller than 1.00 atm. And the total pressure must not change because of the stoichiometry of the reaction: there are 2 moles of the gas reactants for 2 moles of the gas products.
Because is a reversible reaction, it will not go to completion, it will reach an equilibrium, and as discussed above, the partial pressures will change.
Answer:
Explanation:
The pressure, the volume and the temperature of an ideal gas are related to each other by the equation of state:
where
p is the pressure of the gas
V is the volume of the gas
n is the number of moles
R is the gas constant
T is the absolute temperature
For the gas in this problem:
n = 2.00 mol is the number of moles
V = 17.4 L is the gas volume
p = 3.00 atm is the gas pressure
is the absolute temperature
Solving for R, we find the gas constant:
Answer:
Vapour pressure of solution is 78.151 torr
Explanation:
Molar mass of biphenyl = 154.21 g
Molar mass of benzene = 78.11 g
19.2 g biphenyl = (19.2/154.21) moles of biphenyl = 0.125 moles of biphenyl
33.7 g of benzene = (33.7/78.11) moles of benzene = 0.431 moles of benzene
Total number of moles = (0.125+0.431) moles = 0.556 moles
Mole fraction of benzene in solution = (0.431/0.556) = 0.775
According to Roults law, vapour pressure of solution made from non-volatile solute = 
Here solute is biphenyl and solvent is benzene
So, vapour pressure of solution = 
= 78.151 torr
1. Given the following equation: N2 (g) + 3 H2 (g) ↔ 2NH3 (g) ΔH = -92 kJ/mol
a. this reaction is exothermic as ΔH is -ve
b. the equilibrium will shift 2 the left if nitrogen gas is removed
c. the equilibrium shift 2 the right if the temperature is lowered
d. the equilibrium shift 2 the left if ammonia (NH3) is added
e. principle of thermodynamic potential or Gibbs energy is used to answer B-D
