Reaction Rates and Equilibrium: A Comprehensive Chemistry Homework

This chemistry assignment solution covers various aspects of chemical kinetics, equilibrium, and data analysis. It includes calculations related to first-order reactions, such as determining the equation for ln(c) and relating it to the form y=mx+c. The solution also demonstrates how to calculate the ma...Read More

MATH
Question 2:
The rate order for the first order chemical reaction is given by:c=c0 e−kt , where c is the
concentration at time t, c0is the concentration at time t=0
a) Equation for ln(c)
ln ( c )=ln (c0 e−kt )¿ ln ( c0 ) + ln ( e−kt )¿ ln ( c0 ) −kt
b) The equation ln ( c ) =−kt+ ln ( c0 ), is in the form of y=mx+c with slope of −kt and y-
intercept is ln ( c0 ),
Question 3:
Mass required to make 250 cm3, of 0.100 M MgCl2 solution: Molar mass of MgCl2 =95.2
Molarity= moles
Volume 0.100= n
0.25 n=0.1× 0.25=0.025 moles of MgC l2
Mass=Molar mass ×number of moles ¿ 95.2 ×0.025¿ 2.38 g
Question 4
Diluting 25cm3 of 5.00 NaOH with 125 cm3of water. The final volume will be 25+125 =150cm3.
c2 V 2=c1 V 1s c2= c1 V 1
V 2
= 5.00× 25
150 ¿ 0.8333 M NaOH
Question 5
The variation of the equilibrium constant, K with temperature, T for an enzyme reactor

a) Completing the table
T/K 289.0 294.2 298.0 304.9 310.5
K/107 7.25 5.25 4.17 2.66 2.00
103/(T/K) 3.4602 3.3990 3.3557 3.2798 3.2206
ln (K) 18.099
1
17.7763 17.5460 17.0964 16.8112
b) The figure below shows the graph of ln (k) verses 1/T, with the line of best fit
c) The slope is
∆ ( ln ( K ) )
∆ ( 1
T ) =5445

Question 1
From the following data of enthalpy value
∆ H /kJmol−1-28.3 -28.7 -31.2 -33.1 -34.4 -37.2 -39.1
a) Mean
μ= 1
n ∑ xi¿ 1
7 × ( −231.8 ) =−33.11428571 kJmo l−1
b) Median is −33.1 kJmo l−1
c) Range ¿ max−min=−28.3− (−39.1 ) =10.8 kJmol−1
d) Standard deviation
σ = √ ∑ ( Xi−μ ) 2
N −1 = √ 100.0285714
7−1 =4.083066075kJmo l−1
e) Variance
Variance=σ2=4.0830660752=16.67142857 kJmo l−1
Question 2
The data below is the measurements of the concentration of arsenic in ppb
19.7 20.5 29.9 21.2 22.1 23.6
A. The mean of arsenic concentration
μ= 1
n ∑ xi¿ 1
6 × ( 128 ) =21.333333333
B. The standard deviation
σ = √ ∑ ( Xi−μ ) 2
N −1 = √ 9.2933333
6−1 =1.363329258

C. Standard error of the mean
σ M = σ
√n = 1.363329258
√6 =0.556576839
D. 95 confident interval of the mean
μ−t0.95 σ M ≤ X ≤ μ−t0.95 σM
21.333333333−2.57 ( 0.556576839 ) ≤ X ≤21.333333333−2.57 ( 0.556576839 )
19.90293086 ≤ X ≤ 22.76373581
Question 3
V total=V final−V initial ¿ [ 43.20 ± 0.05 ] − [ 25.60 ± 0.05 ] ¿ ( 43.20−25.60 ) ± ( 0.05+0.05 )¿ 17.6 ± 0.10
UV =± 0.10
Question 4
Average speed= Distance
Time ¿ 100± 0.1
10.20± 0.02
m
s ¿ 100
10.20 ± ( 0.1+0.02 ) ¿ 9.80 ± 0.12U speed=± 0.12
Question 5
The result of a fluorescence spectrometry experiment with intensity, I as a function of
concentration
I 2.1 5.0 9.0 12.6 17.3 21.1
C 0.00 2.00 4.00 6.00 8.00 10.00
a) To find the best line of fit is given by I=a+bC, where a is the slope and b is the y- intercept

C I CI C^2 I^2
0 2.1 0 0 4.41
2 5 10 4 25
4 9 36 16 81
6 12.6 75.6 36 158.76
8 17.3 138.4 64 299.29
10 21.1 211 100 445.21
SUM 30 67.1 471 220 1013.67
a=∑ I .∑ C2 −∑ C ∑ IC
N .∑ C2− (∑ C )2 =67.1 ( 220 ) −30(471)
6 ( 220 )−302 =1.50 47619047619
b= n .∑ CI −∑ C ∑ I
n. ∑ C2− ( ∑ C )
2 = 6 ( 471 ) −30(67.1)
6 ( 220 ) −302 =1.93 57142857143
∴ Best line of fit : I=1.9357142857143 C+1.5047619047619
b) Correlation, r
r = n .∑ CI −∑ C ∑ I
√ (n∑ C2 − (∑ C )2
)× ( n∑ I2− (∑ I )2
) ¿ 6 ( 471 ) −30 ( 67.1 )
√ ( 6× 220−302 ) × ( 6 ×1013.67−67.12 )
¿ 0.998138858