Calculate the standard entropy change for the industrial synthesis of urea (a common fertilizer): co2(g) + 2 nh3(g) → co(nh2)2(s) + h2o(ℓ) at 25◦c. patel (smp4358) – 28: fall exam review part three – lyssy – (112476) 8 s ◦ co2(g) 213.74 j k·mol nh3(g) 192.45 j k·mol co(nh2)2(s) 104.6 j k·mol h2o(ℓ) 69.91 j k·mol answer in units of j k · mol.

Answer : The standard entropy change is, -424.1 J/K

Explanation :

The given balanced reaction is,

The expression used for entropy change of reaction is:

where,

= entropy change of reaction = ?

n = number of moles

= standard entropy change

= 104.6 J/mol.K

= 69.91 J/mol.K

= 213.74 J/mol.K

= 192.45 J/mol.K

Now put all the given values in this expression, we get:

Therefore, the standard entropy change is, -424.1 J/K

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g