shows Frenkel defect because of the small size of ion, it can easily fit in the interstitial sites while can not as it is bigger in size.

A binary number uses only two digits zero and one. In a binary system, numbers are represented as follows

NUMBER AS INTEGER | NUMBER IN BINARY |

0 | 00 |

1 | 01 |

2 | 10 |

3 | 11 |

4 | 100 |

5 | 101 |

6 | 110 |

7 | 111 |

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test> <sc

Given function is
Given function is satisfies for the all real values of x
case (i) k < 0
Hence, function is continuous for all values of x < 0
case (ii) x = 0
L.H.L at x= 0
R.H.L. at x = 0
L.H.L. = R.H.L. = f(0)
Hence, function is continuous at x = 0
case (iii) k > 0
Hence , function is continuous for all values of x > 0
case (iv) k < 1
Hence , function is...

From the Arrhenius equation,
...................................(i)
it is given that
T1= 293 K
T2 = 313 K
Putting all these values in equation (i) we get,
Activation Energy = 52.86 KJ/mol
This is the required activation energy

We know that,
for a first order reaction-
Case 1
At temp. = 298 K
= 0.1054/k
Case 2
At temp = 308 K
= 2.2877/k'
As per the question
K'/K = 2.7296
From Arrhenius equation,
= 76640.096 J /mol
=76.64 KJ/mol
k at 318 K
we have , T =318K
A=
Now
After putting the calue of given variable, we get
on takingantilog we get,
k =...

The decomposition of A into a product has a value of k as at 10°C and energy of activation 60 kJ mol–1.
K1 =
K2 =
= 60 kJ mol–1
K2 =

The Arrhenius equation is given by
taking log on both sides,
....................(i)
given equation,
.....................(ii)
On comparing both equation we get,
activation energy
half life () = 256 min
k = 0.693/256
With the help of equation (ii),
T =
= 669 (approx)

The Arrhenius equation is given by
.................................(i)
given equation,
............................(ii)
by comparing equation (i) & (ii) we get,
A= 4.51011 per sec
Activation energy = 28000 (R = 8.314)
= 232.792 KJ/mol

For first order reaction,
given that half life = 3 hrs ()
Therefore k = 0.693/half-life
= 0.231 per hour
Now,
= antilog (0.8024)
= 6.3445
(approx)
Therefore fraction of sample of sucrose remains after 8 hrs is 0.157

Given that,
k =
t = 100 s
Here the unit of k is in per sec, it means it is a first-order reaction.
therefore,
Hence the concentration of rest test sample is 0.135 mol/L

Given that,
k =
= 179.9 KJ/mol
T(temp) = 546K
According to Arrhenius equation,
taking log on both sides,
= (0.3835 - 5) + 17.2082
= 12.5917
Thus A = antilog (12.5917)
A = 3.9 per sec (approx)

From the above data,
T/
0
20
40
60
80
T/K
273
293
313
333
353
()
3.66
3.41
3.19
3.0
2.83
0.0787
1.70
25.7
178
2140
-7.147
-4.075
-1.359
-0.577
3.063
Slope of line =
According to Arrhenius equations,
Slope =
12.30 8.314
= 102.27
Again,
When T = 30 +273 = 303 K and 1/T =0.0033K
k =
When T = 50 + 273 = 323 K and 1/T = 3.1 K
k = 0.607 per sec

The thermal decomposition of is shown here;
After t time, the total pressure =
So,
thus,
for first order reaction,
now putting the values of pressures, when t = 100s
when
= 0.65 - 0.5
= 0.15 atm
So,
= 0.5 - 0.15
= 0.35 atm
Thus, rate of reaction, when the...

Decompostion is represented by equation-
After t time, the total pressure =
So,
thus,
for first order reaction,
now putting the values of pressures,
when t =360sec
when t = 270sec
So,

For the first-order reaction,
(30% already decomposed and remaining is 70%)
therefore half life = 0.693/k
=
= 77.7 (approx)

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