The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation
R = molar gas constant
K = A(e^(-Ea/RT))
Taking natural log of both sides
In K = In A - (Ea/RT)
In K = (-Ea/R)(1/T) + In A
Comparing this to the equation of a straight line; y = mx + c
y = In K, slope, m = (-Ea/R), x = (1/T) and intercept, c = In A
a) From the question, m = (-Ea/R) = -1.10 × (10^4) K
(-Ea/R) = -1.10 × (10^4) = -11000
R = 8.314 J/K.mol
Ea = -11000 × 8.314 = 91454 J/mol = 91.454 KJ/mol
b) c = In A = 33.5
A = e^33.5 = (3.54 × (10^14))/s
c) K = A(e^(-Ea/RT))
A = (3.54 × (10^14))/s, Ea = 91454 J/mol, T = 25°C = 298.15 K, R = 8.314 J/K.mol
K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s
QED!
Answer:
The products are Calcium oxide and Carbon dioxide.
Explanation:
When calcium carbonate is heated, thermal decomposition occurs.
Calcium calcium → Calcium oxide + Carbon dioxide
Answer:
Theoretical yield of the reaction = 34 g
Excess reactant is hydrogen
Limiting reactant is nitrogen
Explanation:
Given there is 100 g of nitrogen and 100 g of hydrogen
Number of moles of nitrogen = 100 ÷ 28 = 3·57
Number of moles of hydrogen = 100 ÷ 2 = 50
Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation
N2 + 3H2 → 2NH3
From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed
Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen
⇒ We require 10·71 moles of hydrogen
But we have 50 moles of hydrogen
∴ Limiting reactant is nitrogen and excess reactant is hydrogen
From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia
Molecular weight of ammonia = 17 g
∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g
Answer:
bdndbdjdbdjdbdjdbsidbdidbsjsbsisbsidbd
23 pairs of chromosomes, since there is 46 singular chromosomes.
