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The mass of N2O4 in the final equilibrium mixture is 39.45 grams.
The decomposition reaction of N2O4 to 2NO2 can be expressed as:
From the parameters given:
The number of mole of N2O4 can be determined as:
From stochiometry, the concentration of
The I.C.E table can be computed as:
↔
Initial 0.2717 0
Change -x +2x
Equilibrium (0.2717 - x) 2x
The equilibrium constant from the I.C.E table can be expressed as:
∴
0.0361 - 0.133x = 4x²
4x² + 0.133x - 0.0361 = 0
By solving the above quadratic equation, we have;
x = 0.07978
The Concentration of [NO2] = 2x
The Concentration of [N2O4] = 0.2717 - x
Again, from the decomposition reaction, we can assert that;
0.19192 M of N2O4 decompose to produce 0.15956 M of NO2.
However, if 5.0 g of NO2 is injected into the vessel, then the number of moles of NO2 injected becomes;
The Molarity of NO2 injected now becomes:
So, the new moles of [NO2] becomes = 0.15956 + 0.05434
= 0.2139 M
The new I.C.E table can be computed as:
↔
Initial 0.19192 0.2139 M
Change +x -2x
Equilibrium (0.19192 + x) (0.2139 -2x)
NOTE: The injection of NO2 makes the reaction proceed in the backward direction.
The equilibrium constant from the I.C.E table can be expressed as:
By solving for x;
x = 0.2246 or x = 0.0225
We need to consider the value of x that is less than the initial concentration of NO2(0.2139 M) which is:
x = 0.0225
Now, the final concentration of [N2O4] = (0.19192 + 0.0225)M
= 0.21442 M
The final number of moles of N2O4 = Molarity(concentration) × volume
The final number of moles of N2O4 = (0.21442 × 2) moles
The final number of moles of N2O4 = 0.42884 moles
The mass of N2O4 in the final equilibrium mixture is:
= final number of moles × molar mass of N2O4
= 0.42884 moles × 92 g/mol
= 39.45 grams
Learn more about the decomposition of N2O4 here: