NAMA : LAURA ELVIRANDRA NIM : 320200404014
PRODI : TEKNIK KONSTRUKSI BANGUNAN MILITER
1.
Answer:
Known : The reaction is 3 / 2 order in CH3CHO.
Qustion : Plot ln k versus 1/T and determine the activation energy!
ln k =( −Ea R ¿(1
T)+lnA
ln k 1/T
-4,51 1,43 x 10-3
-3,35 1,37 x 10-3
-2,254 1,32 x 10-3
-1,070 1,27 x 10-3
-0,237 1,23 x 10-3
A plot of these data yields the graph in figure, the slope of the line is calculated from two pairs
of coordinates :
Slope = −4.00−(−0.45)
(1.41−1.24)10−3 = -2.09 x 104 K Slope = −Ea
R = -2.04 x 104 K Ea = (8,314J/K.mol) (2.09x104 K) Ea = 1,74 x 102kJ/mol
So, the energy activation is 1,74 x 102 kJ/mol
2.
Answer:
When there is an increase in pressure, the equilibrium will shift towards the side of the reaction with fewer moles of gas.
When there is a decrease in pressure, the equilibrium will shift towards the side of the reaction with more moles of gas.
So we can conclude a.
Based on the reaction, it can be seen that the number of coefficients of the reactants (2 + 3 = 5) is more than the coefficients of the products (2 + 2 = 4). So with the theory above, we can conclude that the direction is right or toward the product
b.
Based on thereaction, it can be seen that the number of coefficients of the products (1+1 = 2) is more than the coefficients of the reactant (1). So, we can conclude that the direction is left or toward the reactan.
c.
Based on the reaction, it can be seen that the number of coefficients of the products (1+1 = 2) is equals value coefficients of the reactant (1+1=2). So with the theory above, we can conclude that the reaction already equilibrium
