Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.110 atm to rise
to 0.150 atm .
1 answer:
Answer:
t = 37.1 s
Explanation:
The equation for the reaction is given as;
2 N2O5(g) --> 4 NO2 + O2
Initial: 0.110 - -
change: -2x +4x +x
Final: 0.110 - 2x +4x +x
But final = 0.150atm;
0.110 - 2x + 4x + x = 0.150 atm
3x = 0.150 - 0.110
x = 0.0133 atm
Pressure in reactant side;
0.110 - 2x
0.110 - 2 (0.0133) = 0.0834 atm
The integral rate law expression is given as;
ln ( [A] / [Ao] ) = -kt
k = rate constant = 7.48*10^-3*s-1
ln (0.0834/0.11) = (7.48*10^-3) t
upon solving, t = 37.1 s
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