Answer:
4600s
Explanation:
For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:
![-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D%3Dk%5BB%5D%20-%20-%20-%20%20-%5Cfrac%7Bd%5BB%5D%7D%7B%5BB%5D%7D%3Dk%2Adt)
If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.
PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.
![-\frac{d[P(B)]}{P(B)}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BP%28B%29%5D%7D%7BP%28B%29%7D%3Dk%2Adt)
![-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%7D%7BP%28N_%7B2%7DO_%7B5%7D%29%7D%3Dk%2Adt)
Integrating we get:
![\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ep%20%5C%2C-%5Cfrac%7Bd%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%7D%7BP%28N_%7B2%7DO_%7B5%7D%29%7D%3D%5Cint%5Climits%5E%20t%20k%2Adt)
![-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})](https://tex.z-dn.net/?f=-%28ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D-ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%29%5D%29%3Dk%28t_%7B2%7D-t_%7B1%7D%29)
Clearing for t2:
![\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}](https://tex.z-dn.net/?f=%5Cfrac%7B-%28ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D-ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%29%5D%29%7D%7Bk%7D%2Bt_%7B1%7D%3Dt_%7B2%7D)
![ln[P(N_{2}O_{5})]=ln(650)=6.4769](https://tex.z-dn.net/?f=ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%3Dln%28650%29%3D6.4769)
![ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333](https://tex.z-dn.net/?f=ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%5D%3Dln%28760%29%3D6.6333)
Answer:
2.803013439419911 × 10⁻¹² J
Explanation:
Mass defect = mass of reactant - mass of product
(2.0140 + 3.01605) - (4.002603 + 1.008665)
5.03005 - 5.011268 = 0.018782 amu
mass in Kg = mass (amu) × 1.66053892173 × 10⁻²⁷ kg
mass in kg = 0.018782 × 1.66053892173 × 10⁻²⁷ = 3.1188242027932 × 10⁻²⁹kg
E = Δm c² where c is the speed of light = 2.9979 × 10⁸m/s
E = 3.1188242027932 × 10⁻²⁹kg × (2.9979 × 10⁸m/s)² = 2.803013439419911 × 10⁻¹² J
Answer:
New temperature T2 = 707.5 K (Approx.)
Explanation:
Given:
Old pressure P1 = 2 atm
Old temperature T1 = 283 K
New Pressure P2 = 5 atm
Find:
New temperature T2
Computation:
Using Gay-Lussac law;
P1 / T1 = P2 / T2
So,
2 / 283 = 5 / T2
New temperature T2 = 707.5 K (Approx.)
no...the atoms will not behave the same
as when temperature is increased, the atoms vibration and kinetic energy will also be increased....they come in excited state...
where as when temperature is reduced ,atoms kinetic energy slows down....
D i think if not here’s this a collection of beliefs or practices mistakenly regarded as being based on scientific method
