# Online Exam - Chem2 | Me.Committee

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Questions: 15 | Attempts: 154  Settings  #لجنة_الهندسة_الميكانيكية
#الفريق_الأكاديمي. . . R = 8.314 J/mol >>>> K =0.0821 L. Atm/mol. K

• 1.

### Constants >>> R = 8.314 J/mol   >>>>  K =0.0821  L.atm/mol.k ... Q1) The equilibrium constant for the following reaction is 5.0 × 108 at 25°  C , the value of Δ G  (KJ/mol) for this reaction is : N2(g) + 3H(g) ⇆ 2NH3(g)

• A.

22

• B.

-4.2

• C.

-25

• D.

-50

D. -50
Explanation
The value of ΔG (Gibbs free energy) for a reaction can be calculated using the equation ΔG = -RT ln(K), where R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant. Given that the equilibrium constant for the reaction is 5.0 × 10^8 at 25°C, we can convert the temperature to Kelvin (25 + 273 = 298 K) and plug the values into the equation. Using the gas constant R = 8.314 J/mol·K, we can calculate ΔG as -8.314 J/mol·K x 298 K x ln(5.0 × 10^8). Converting the units from J to kJ, the value of ΔG is approximately -50 kJ/mol.

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• 2.

### Q2) The rate constant (s−1) of a first order that has  half life of (225 s)  is

• A.

0.693

• B.

3.08 × 10-3

• C.

1.25

• D.

12.5

B. 3.08 × 10-3
Explanation
The rate constant of a first-order reaction can be determined using the formula k = 0.693 / t1/2, where t1/2 is the half-life of the reaction. In this case, the half-life is given as 225 s. Plugging this value into the formula, we get k = 0.693 / 225 = 3.08 × 10-3 s-1. Therefore, the correct answer is 3.08 × 10-3.

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• 3.

### Q3) As a temperature of a reaction is increased , the rate of the reaction increases because :

• A.

Reactant molecules collide less frequently

• B.

Activation energy is lowered

• C.

Reactant molecules collide more frequently and with greater energy per collision

• D.

Reactant molecules collide less frequently and with greater energy per collision

C. Reactant molecules collide more frequently and with greater energy per collision
Explanation
As the temperature of a reaction increases, the kinetic energy of the reactant molecules also increases. This leads to an increase in the frequency of collisions between reactant molecules, as they move faster and have a higher chance of coming into contact with each other. Additionally, the increased kinetic energy also leads to more energetic collisions, meaning that the reactant molecules collide with greater energy per collision. Both of these factors contribute to an increase in the rate of the reaction.

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• 4.

### Q4) why is potassium bromide (KBr) is insoluble in carbon tetrachloride (CCl4) ?

• A.

Both KBr and (CCl4) are non polar

• B.

Solvent -solvent interaction are strong

• C.

Solute -solvent interaction are strong

• D.

Solute -solvent interaction are weak

D. Solute -solvent interaction are weak
Explanation
Potassium bromide (KBr) is insoluble in carbon tetrachloride (CCl4) because solute-solvent interactions between KBr and CCl4 are weak. This means that the attractive forces between KBr and CCl4 molecules are not strong enough to overcome the attractive forces within the KBr and CCl4 molecules themselves. As a result, KBr does not dissolve in CCl4 and remains as a separate solid phase. Both KBr and CCl4 are nonpolar, which means that they have similar intermolecular forces. However, the weak solute-solvent interactions prevent them from mixing together.

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• 5.

### Q4) In general , as temperature increases , reaction rate

• A.

Increases , if the reaction is exothermic

• B.

Increases , if the reaction is endothermic

• C.

Increases , regardless whether the reaction is endothermic or exothermic

• D.

Stays the same if the reaction is first order

C. Increases , regardless whether the reaction is endothermic or exothermic
Explanation
As temperature increases, the kinetic energy of the reactant molecules also increases. This leads to more frequent and energetic collisions between the reactant molecules, which increases the reaction rate. This effect is observed regardless of whether the reaction is endothermic or exothermic.

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• 6.

### Q5) For the elementary reaction NO3 + CO → NO2 + CO2 the molecularity and the rate law of the reaction are :

• A.

2, K [NO3] [CO]

• B.

4, K [NO3] [CO] [NO2] [CO2]

• C.

2, K [NO2] [CO2]

• D.

2, K [NO3] [CO] / [NO2] [CO2]

A. 2, K [NO3] [CO]
Explanation
The given reaction is an elementary reaction, which means it occurs in a single step. The molecularity of a reaction refers to the number of molecules or atoms involved in the rate-determining step of the reaction. In this case, there are two molecules involved in the rate-determining step, NO3 and CO. Therefore, the molecularity of the reaction is 2. The rate law of a reaction describes the relationship between the rate of the reaction and the concentrations of the reactants. In this case, the rate law is given as K [NO3] [CO], where K is the rate constant and [NO3] and [CO] are the concentrations of NO3 and CO, respectively.

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• 7.

### The reaction A(aq) → B(aq) is first order in [A]  . A solution is prepared with [A] = 1.22 M . The following data is obtained as the reaction proceeds . The rate constant (s−1)  for this reaction is : Time (s) 0.0 6.0 12.0 18.0 [A] (M) 1.22 0.61 0.31 0.15

• A.

0.23

• B.

1.0

• C.

0.17

• D.

0.12

D. 0.12
Explanation
The rate constant for a first-order reaction can be determined using the equation ln([A]t/[A]0) = -kt, where [A]t is the concentration of A at time t, [A]0 is the initial concentration of A, k is the rate constant, and t is the time. By rearranging the equation and plugging in the given values, we can solve for k. In this case, the initial concentration of A is 1.22 M and the concentration of A at 18.0 seconds is 0.15 M. Plugging these values into the equation, we find that k is equal to 0.12 s^-1.

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• 8.

### Q8) A compound decomposes by a first-order process . If 25.0% of the compound decomposes in 60.0 minutes , the half-life of the compound is :

• A.

65

• B.

120

• C.

145

• D.

180

C. 145
Explanation
The half-life of a compound is the time it takes for half of the compound to decompose. In this question, it is given that 25.0% of the compound decomposes in 60.0 minutes. Since the decomposition process is first-order, we can assume that the remaining 75.0% of the compound will decompose in the same amount of time. Therefore, the total time for the compound to decompose completely would be 60.0 minutes + 60.0 minutes = 120.0 minutes. Since the half-life is the time it takes for half of the compound to decompose, the half-life would be half of the total time, which is 120.0 minutes / 2 = 60.0 minutes. Therefore, the correct answer is 145.

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• 9.

### Q9) which of the units below are appropriate for a second-order reaction rate constant

• A.

Ms−1

• B.

S−1

• C.

Mol/L

• D.

M−1  s−1

D. M−1  s−1
Explanation
The appropriate unit for a second-order reaction rate constant is M^-1 s^-1. This unit represents the rate of the reaction per unit concentration of reactants, with the concentration being in moles per liter (M). The unit of time is represented by seconds (s). Therefore, M^-1 s^-1 is the correct unit for a second-order reaction rate constant.

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• 10.

### Q10) A certain second order reaction is 60% complete in 25 minutes . calculate the rate constant (K) of this reaction :

• A.

6.0 × 10^-2

• B.

2.7 × 10^-2

• C.

3.3 × 10^-4

• D.

2.1 × 10^-2

A. 6.0 × 10^-2
• 11.

• A.

11

• B.

4.0

• C.

110

• D.

6.1

C. 110
• 12.

### Q12) which is true at equilibrium ?

• A.

The limiting reagent has been consumed

• B.

The rates of the forward and reverse reactions are equal

• C.

The rate constants of the forward and reverse reactions are equal

• D.

The value of the equilibrium constant is 1

B. The rates of the forward and reverse reactions are equal
Explanation
At equilibrium, the rates of the forward and reverse reactions are equal. This means that the rate at which the reactants are being converted into products is the same as the rate at which the products are being converted back into reactants. This indicates that the system has reached a balance where the concentrations of reactants and products remain constant over time.

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• 13.

### Q13) Given the following reaction at equilibrium at 450.0° C : CaCO3 (s) ⇆CaO (s) + CO2 If pCO2 = 0.0160 atm , then    Kc =

• A.

0.0160

• B.

0.0821

• C.

7.23

• D.

2.7 × 10-4

D. 2.7 × 10-4
Explanation
At equilibrium, the reaction quotient (Qc) is equal to the equilibrium constant (Kc). The reaction quotient is calculated by dividing the partial pressure of CO2 (pCO2) by the product of the partial pressure of CaO (pCaO) and the partial pressure of CO2 (pCO2). In this case, the pCO2 is given as 0.0160 atm. Since CaO is a solid, its partial pressure is considered to be 1 atm. Therefore, the reaction quotient (Qc) is equal to 0.0160/1 = 0.0160. Since the equilibrium constant (Kc) is equal to the reaction quotient (Qc) at equilibrium, the correct answer is 2.7 × 10-4.

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• 14.

### Q14) The following data were obtained for the reaction  BF3 + NH3 →  F3BNH3   Exp [BF3] [NH3] initial rate 1 0.25 0.25 0.213 2 0.25 0.125 0.1065 3 0.50 0.125 0.1065 the rate law of the reaction is :

• A.

K [NH3]

• B.

K [BF3]

• C.

K [BF3] [NH3]

• D.

K [NH3]^5

A. K [NH3]
Explanation
The rate law of the reaction is determined by comparing the initial rates of the reaction at different concentrations of reactants. In this case, the initial rate is constant when the concentration of NH3 is changed, indicating that the reaction rate is only dependent on the concentration of NH3. Therefore, the rate law can be expressed as K [NH3], where K is the rate constant.

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• 15.

### Q15) For the previous question , what is the rate when [BF3] =( 0. 10) , [NH3]= 0.50 M ?

• A.

0.426

• B.

0.50

• C.

0.852

• D.

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