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The independent variable is the variable that you have control of. By changing this variable, the other dependent variables will consequently change. Thus, the independent variable nature of the mercury oxide sample used.
Answer:
a) First-order.
b) 0.013 min⁻¹
c) 53.3 min.
d) 0.0142M
Explanation:
Hello,
In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.
a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.
b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:
c) Half life for first-order kinetics is computed by:
d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:
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Answer:
∆G°= -55005J or -55KJ
Explanation:
The first step is to determine E°cell= E°cathode - E°anode
2Cl-(aq)/Cl2(g) system is the cathode while 2Br-(aq)/Br2(g) is the anode
E°cell= 1.360-1.075
E°cell= 0.285V
From
∆G°= -nFE°cell
n= 2 from the balanced reaction equation, two electrons were transferred.
F= 96500
E°cell=0.285V
∆G°= -(2×96500×0.285)
∆G°= -55005J or -55KJ
Answer:
Difference of the enthalpy (of a system) minus the product of the entropy and absolute temperature
Explanation:
The basis of spontaneity in a chemical reaction is that ∆G must be negative. ¡∆G is known as the change in free energy of a system. If ∆G is negative, then the reaction will occur without any external help (the reaction is spontaneous at room temperature).
∆G is given by;
∆G= ∆H -T∆S
Where;
∆H= change in enthalpy of the system
T= absolute temperature of the system
∆S= change in entropy
Hence; when ∆H -T∆S gives a negative result, the reaction proceeds without any external help.