Answer:
76.03 °C.
Explanation:
Equation:
C2H5OH(l) --> C2H5OH(g)
ΔHvaporization = ΔH(products) - ΔH (reactants)
= (-235.1 kJ/mol) - (-277.7 kK/mol)
= 42.6 kJ/mol.
ΔSvaporization = ΔS(products) - ΔS(reactants)
= 282.6 J/K.mol - 160.6 J/K.mol
= 122 J/K.mol
= 0.122 kJ/K.mol
Using gibbs free energy equation,
ΔG = ΔH - TΔS
ΔG = 0,
T = ΔH/ΔS
T = 42.6/0.122
= 349.18 K.
Coverting Kelvin to °C,
= 349.18 - 273.15
= 76.03 °C.
Those are both correct! great job, keep up the good work (-:
Answer:
0.17325 moles per liter per second
Explanation:
For a first order reaction;
in[A] = in[A]o - kt
Where;
[A]= concentration at time t
[A]o = initial concentration
k= rate constant
t= time taken
ln0.5 =ln1 - 2k
2k = ln1 - ln0.5
k= ln1 - ln0.5/2
k= 0 -(0.693)/2
k= 0.693/2
k= 0.3465 s-1
Rate of reaction = k[A]
Rate = 0.3465 s-1 × 0.50 mol/L
Rate = 0.17325 moles per liter per second
Answer:
%age Yield = 51.45 %
Solution:
Step 1: Convert Kg into g
68.5 Kg CO = 68500 g CO
8.60 Kg H₂ = 8600 g
Step 2: Find out Limiting reactant;
The Balance Chemical Equation is as follow;
CO + 2 H₂ → CH₃OH
According to Equation,
28 g (1 mol) CO reacts with = 4 g (2 mol) of H₂
So,
68500 g CO will react with = X g of H₂
Solving for X,
X = (68500 g × 4 g) ÷ 28 g
X = 9785 g of H₂
It shows 9785 g H₂ is required to react with 68500 g of CO but we are provided with 8600 g of H₂ which is less than required. Therefore, H₂ is provided in less amount hence, it is a Limiting reagent and will control the yield of products.
Step 3: Calculate Theoretical Yield
According to equation,
4 g (2 mol) H₂ reacts to produce = 32 g (1 mol) Methanol
So,
8600 g H₂ will produce = X g of CH₃OH
Solving for X,
X = (8600 g × 32 g) ÷ 4 g
X = 68800 g of CH₃OH
Step 4: Calculate %age Yield
%age Yield = Actual Yield ÷ Theoretical Yield × 100
Putting Values,
%age Yield = 3.54 × 10⁴ g ÷ 68800 g × 100
%age Yield = 51.45 %
