Answer:
The rate law for second order unimolecular irreversible reaction is
![\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%20k.t%20%2B%20%5Cfrac%7B1%7D%7B%5BA%5D_%7B0%7D%20%7D)
Explanation:
A second order unimolecular irreversible reaction is
2A → B
Thus the rate of the reaction is
![v = -\frac{1}{2}.\frac{d[A]}{dt} = k.[A]^{2}](https://tex.z-dn.net/?f=v%20%3D%20-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3D%20k.%5BA%5D%5E%7B2%7D)
rearranging the ecuation
![-\frac{1}{2}.\frac{k}{dt} = \frac{[A]^{2}}{d[A]}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bk%7D%7Bdt%7D%20%3D%20%5Cfrac%7B%5BA%5D%5E%7B2%7D%7D%7Bd%5BA%5D%7D)
Integrating between times 0 to <em>t </em>and between the concentrations of ![[A]_{0}](https://tex.z-dn.net/?f=%5BA%5D_%7B0%7D)
to <em>[A].</em>
![\int\limits^0_t -\frac{1}{2}.\frac{k}{dt} =\int\limits^A_{0} _A\frac{[A]^{2}}{d[A]}](https://tex.z-dn.net/?f=%5Cint%5Climits%5E0_t%20-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bk%7D%7Bdt%7D%20%3D%5Cint%5Climits%5EA_%7B0%7D%20_A%5Cfrac%7B%5BA%5D%5E%7B2%7D%7D%7Bd%5BA%5D%7D)
Solving the integral
Answer:
ΔT = Tfinal − Tinitial = 150°C − 35.0°C = 125°C
given the specific heat of iron as 0.108 cal/g·°C
heat=(100.0 g)(0.108 cal /g· °C )(125°C) =
100x 0.108x125= 1350 cal
A) Cu
Cu + 2HCl --> CuCl2 + H2(g)
Products predicted: Copper(II) choloride and hydrogen gas
B) Mg
Mg + 2HCl --> MgCl2 + H2
Products predicted: magnesium chloride + hygrogen gas
C) Fe
Fe +2 HCl -> FeCl2 + H2, or
2Fe +6 HCl -> 2FeCl3 + 3H2
Products predicted: Iron(II) chloride, iron (III) chloride and hydrogen gas.
Heterogenous mixtures are unevenly mixed. Like oil and vinegar in vinaigrette if it is not emulsified well enough and they separate. Any case where two things are not evenly distributed within each other.
Homogenous mixtures are evenly mixed throughout. Like salt water or kool-aid (when it's mixed).
Hope this helps!
Heat produced = -13588.956 kJ
<h3>Further explanation</h3>
Given
The reaction of combustion of Methane
CH4(g)+2O2(g)→CO2(g)+2H2O(g) ΔH∘rxn=−802.3kJ
271 g of CH4
Required
Heat produced
Solution
mol of 271 g CH₄ (MW=16 g/mol0
mol = mass : MW
mol = 271 : 16
mol = 16.9375
So Heat produced :
= mol x ΔH°rxn
= 16.9375 mol x −802.3kJ/mol = -13588.956 kJ
