The half life of a first order decomposition of NH4NO3 is 2.10hr. If 6...
The half life of a first order decomposition of NH4NO3 is 2.10hr. If 6...
Explanation:
First Order Decomposition: The rate of decomposition of NH4NO3 is directly proportional to the amount of NH4NO3 present. Therefore, the decomposition of NH4NO3 follows first order kinetics.
Half-Life: The half-life of a first order reaction is the time required for the concentration of the reactant to decrease to half of its initial value.
Given,
Half-life (t1/2) = 2.10 hours
We can use the following equation to calculate the rate constant (k) of the reaction.
t1/2 = (0.693/k)
k = (0.693/t1/2) = (0.693/2.10) = 0.33 hr^-1
Time Required for 90% Decomposition:
We can use the following equation to calculate the time required for 90% decomposition of NH4NO3.
ln (Co/C) = kt
Where, Co = initial concentration of NH4NO3
C = concentration of NH4NO3 at time t
Here, Co = 6.2 g (given)
C = 0.1 Co = 0.62 g (since 90% is decomposed)
Substituting the values, we get:
ln (6.2/0.62) = (0.33 t)
t = 6.978 hours
Therefore, the time required for 90% decomposition of NH4NO3 is 6.978 hours.
Volume of N2O Produced:
NH4NO3 decomposes to form N2O and H2O in the ratio of 1:2. Therefore, the moles of N2O produced can be calculated as follows:
Moles of NH4NO3 = 6.2 g / molar mass of NH4NO3
= 6.2 g / (14 + 4 + 14 + 3*16)
= 0.046 moles
Moles of N2O produced = 0.5 * moles of NH4NO3
= 0.5 * 0.046
= 0.023 moles
Using the ideal gas equation, we can calculate the volume of N2O produced at STP.
PV = nRT
Where, P = pressure = 1 atm
V = volume of N2O produced
n = moles of N2O produced
R = gas constant = 0.0821 L atm / mol K
T = temperature at STP = 273 K
Substituting the values, we get:
V = (nRT)/P
= (0.023 * 0.0821 * 273) / 1
= 1.562 L
Therefore, the volume of N2O produced at STP is 1.562 L.
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