Green plants use light from the Sun to drive photosynthesis, a chemical reaction in which liquid water and carbon dioxide gas fo
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Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The formation reaction of will be,
The intermediate balanced chemical reaction will be,
(1)
(2)
(3)
Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :
(1)
(2)
(3)
The expression for enthalpy of formation of will be,
Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole
Given: Half-life of <span>133Ba = t1/2 = 10.5 years.
The radio-active materials obeys 1st order dissociation kinetics. Therefore we have:
k = 0.693 / t1/2 = 0.693 / 10.5 = 0.066 years-1.
Also, </span>
where, Co = initial concentration = <span>1.1×10^10 atoms
Ct = conc. of Ba at time t.
......................................................................................................................
Answer 1: For t = </span><span>5 years
</span>
Therefore,
= 0.1432
Therefore, Co/Ct = 1.3908
Therefore, Ct = 7.9086 X 10^9 atoms.
Number of 133Ba atoms left after 5 years = 7.9086 X 10^9.
....................................................................................................................
Answer 2: For t = 30 years
Therefore, = 0.8597
Therefore, Co/Ct = 7.2402
Therefore, Ct = 1.5193 X 10^9 atoms.
Number of 133Ba atoms left after 30 years = 1.5193 X 10^9.
........................................................................................................................
Answer 3: t = 180 years
Therefore, = 5.1585
Therefore, Co/Ct = 1.44 X 10^5
Therefore, Ct = 7.6367 X10^4 atoms.
Number of 133Ba atoms left after 180 years = 7.6367 X10^4.
The radio-active materials obeys 1st order dissociation kinetics. Therefore we have:
k = 0.693 / t1/2 = 0.693 / 10.5 = 0.066 years-1.
Also, </span>
where, Co = initial concentration = <span>1.1×10^10 atoms
Ct = conc. of Ba at time t.
......................................................................................................................
Answer 1: For t = </span><span>5 years
</span>
Therefore,
Therefore, Co/Ct = 1.3908
Therefore, Ct = 7.9086 X 10^9 atoms.
Number of 133Ba atoms left after 5 years = 7.9086 X 10^9.
....................................................................................................................
Answer 2: For t = 30 years
Therefore,
Therefore, Co/Ct = 7.2402
Therefore, Ct = 1.5193 X 10^9 atoms.
Number of 133Ba atoms left after 30 years = 1.5193 X 10^9.
........................................................................................................................
Answer 3: t = 180 years
Therefore,
Therefore, Co/Ct = 1.44 X 10^5
Therefore, Ct = 7.6367 X10^4 atoms.
Number of 133Ba atoms left after 180 years = 7.6367 X10^4.