Be sure to answer all parts. δg o for the reaction h2(g) + i2(g) ⇌ 2hi(g) is 2.60 kj/mol at 25°c. calculate δg, and predict the direction in which the reaction is spontaneous. the initial pressures are: ph2 = 3.10 atm pi2 = 1.5 atm phi = 1.75 atm

δg =
a. the reaction is spontaneous in the forward direction
b. the reaction is spontaneous in the reverse direction
c. cannot be determined

∆G = 1.567 kJ/mole

The reaction is nonspontaneous., or will be spontaneous in the reverse direction

Explanation:

Step 1: The balanced equation

H2(g) + I2(g) ⇌ 2HI(g)        2.60 kJ/mol at 25°C.

Step 2: Data given

The initial pressures are:

pH2 = 3.10 atm

pI2 = 1.5 atm

pHI = 1.75 atm

Step 3: Calculate Q

Kp = (pHI)^2 / (pH2)(pI2)

Q has the same form but we substitute the non-equilibrium values given in the problem.

Q = (1.75)² / (3.10*1.5) = 0.659

Step 4: Calculate gibbs free energy

∆G = ∆G˚ + RTlnQ = 2600 J/mole + (8.31 J/mole K)(298 K)(ln 0.659) = 2600 J/mole - 1032.729 J/mole

= 1567.271‬ J/mole = 1.567 kJ/mole

Interpret ∆G:

∆G > 0 the reaction is nonspontaneous., or will be spontaneous in the reverse direction

∆G < 0 the reaction is spontaneous.

If ∆G  = 0 the reaction is essentially at equilibrium.

∆G = 1.567 kJ/mole

The reaction is nonspontaneous., or will be spontaneous in the reverse direction

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"