Difficult Chemistry Questions And Answers

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  • 1/28 Questions

    If 50.0 g of benzene, C6H6, at 25.00C absorbs 2.71 kJ of energy in the form of heat, what is the final temperature of the benzene? The specific heat of benzene is 1.72 J/g·0C.

    • 25.00C
    • 31.50C
    • 56.50C
    • 32.30C
    • 57.30C
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About This Quiz

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Difficult Chemistry Questions And Answers - Quiz

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  • 2. 

    If 495 J is required to change the temperature of 12.7 g of sodium chloride from 75.00C to 135 0C, what is the specific heat of sodium chloride?

    • 0.866 J/g·0C

    • 2.60 J/g·0C

    • 0.650 J/g·0C

    • 1.15 J/g·0C

    • 2.83 × 105 J/g·0C

    Correct Answer
    A. 0.650 J/g·0C
    Explanation
    Solving: q = mass ×c ×Δt
    495 J = 12.7 g ×c× (135 – 75)0C c = 495 J / 12.7g · 60 0C = 0.650 J/g ·0C

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  • 3. 

    Water has a specific heat of 4.18 J/g · 0C. If 35.0 g of water at 98.8 0C loses 4.94 kJ of heat, what is the final temperature of the water?

    • 32.00C

    • 46.20C

    • 47.20C

    • 57.20C

    • 65.0C

    Correct Answer
    A. 65.0C
    Explanation
    Solving: loses heat, so -4.94 kJ = 35.0 g ×4.18 J/ g ·0C × (Tf – 98.8 0C), Tf = 65.0 0C

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  • 4. 

    When 66.0 g of an unknown metal at 28.50C is placed in 83.0 g H2O at 78.50C, the water temperature decreases to 75.90C. What is the specific heat capacity of the metal? The specific heat capacity of water is 4.184 J/g0C.

    • 0.055 J/g· 0C

    • 0.29 J/g · 0C

    • 0.69 J/g · 0C

    • 0.18 J/g · 0C

    • 2.6 J/g · 0C

    Correct Answer
    A. 0.29 J/g · 0C
    Explanation
    Solving: qreceive = -qrelease
    66.0g ×c × (78.5 – 28.5) 0C = -83.0 g ×4.184 J/g0C ×(75.9 – 78.5) 0C,
    c = 83.0 ×4.184 J/g0C× 2.6 / 66.0 × 50 = 0.27 J/g0C

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  • 5. 

    A coffee-cup calorimeter contains 10.0 g of water at 59.000C. If 3.00 g gold at 15.20 0C is placed in the calorimeter, what is the final temperature of the water in the calorimeter? The specific heat of water is 4.18 J/g ·0C; the specific heat of gold is 0.128 J/g ·0C.

    • 55.37 C

    • 58.60 C

    • 59.40 C

    • 60.80 C

    • 64.19 C

    Correct Answer
    A. 58.60 C
    Explanation
    Solving: qreaction = -qcarlorimeter
    3.00 g ×0.128 J/g0C × (Tf – 15.2 0C) = - 10.0 g × 4.18 J/g0C × (Tf –59.000C)
    0.384 Tf – 5.84 = -41.8 Tf +2466.2 42.184 Tf = 2472.04 Tf = 58.6 0C

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  • 6. 

    When 10.0 g KOH is dissolved in 100.0 g of water in a coffee-cup calorimeter, the temperature rises from 25.18 0C to 47.53 0C. What is the enthalpy change per gram of KOH dissolved in the water? Assume that the solution has a specific heat capacity of 4.18 J/g0C.

    • –116 J/g

    • –934 J/g

    • –1.03 × 103 J/g

    • –2.19 × 103 J/g

    • –1.03 × 104 J/g

    Correct Answer
    A. –1.03 × 103 J/g
    Explanation
    Solving: ΔH/g =-(100.0 g+ 10.0g)×4.18 J/g0C ×(47.53 -25.18) 0C /10 .0 g = –1.03 × 103 J/g

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  • 7. 

    10.0 g of ice at 0.00 0C is mixed with 25.0 g of water at 35.000C in a coffee-cup calorimeter. What is the final temperature of the mixture? The specific heat of water is 4.18 J/g 0C; the heat of fusion of water is 333 J/g.

    • 0.00 0C

    • 2.24 0C

    • 5.22 0C

    • 25.0 0C

    • 47.8 0C

    Correct Answer
    A. 2.24 0C
    Explanation
    Solving: 10.0 g× 333 J/g = -25.0 g ×4.18 J/g 0C × (T – 35) 0C, 104.5 T = 3657.5 – 3330, T = 3.13 0C

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  • 8. 

    All of the following statements are true EXCEPT

    • Hess' law states that ΔH for an overall reaction is the sum of the ΔH values for the individual equations.

    • The molar enthalpy of formation of a compound is equal to the enthalpy change when one mole of the compound is formed from elements.

    • A reaction with a negative enthalpy is exothermic.

    • The enthalpy of formation of an element in its most stable state is equal to zero.

    • The sum of the enthalpies of formation of the products in a chemical reaction is defined as the enthalpy of reaction.

    Correct Answer
    A. The sum of the enthalpies of formation of the products in a chemical reaction is defined as the enthalpy of reaction.
    Explanation
    The correct answer is "the sum of the enthalpies of formation of the products in a chemical reaction is defined as the enthalpy of reaction." This statement is not true because the enthalpy of reaction is defined as the difference between the enthalpies of the products and the reactants, not the sum of the enthalpies of formation of the products. The enthalpy of reaction takes into account the stoichiometric coefficients of the balanced chemical equation.

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  • 9. 

    Determine the heat of reaction for the combustion of ammonia,         4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(l) using molar enthalpies of formation. NH3(g)    –45.9 NO2(g)    +33.1 H2O(l)    –285.8

    • +30.24 kJ

    • –206.9 kJ

    • –298.6 kJ

    • –1398.8 kJ

    • –1663.6 kJ

    Correct Answer
    A. –1398.8 kJ
    Explanation
    Solving: ΔH = Σ ΔfH0 products - Σ ΔfH0reactants = 4× 33.1 + 6×(-285.8) - 4×(–45.9) = –1398.8 kJ

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  • 10. 

    The standard molar enthalpy of formation of NH3(g) is –45.9 kJ/mol. What is the enthalpy change if 9.51 g N2(g) and 1.96 g H2(g) react to produce NH3(g)?

    • –10.3 kJ

    • –20.7 kJ

    • –29.8 kJ

    • –43.7 kJ

    • –65.6 kJ

    Correct Answer
    A. –29.8 kJ
    Explanation
    Solving: N2(g) + 3H2(g) → 2NH3(g) 9.51/28 = 0.3396 mol N2(g), 1.96 /2 = 0.98 molH2(g) so H2 is limiting reactant, need 0.3267 mol N2(g) only and produce 0.6534 mol NH3(g). ΔH = 0.6534mol × (–45.9 kJ/mol) = -29.9 kJ

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  • 11. 

    For the reaction below, relate the rate of disappearance of hydrogen to the rate of formation of ammonia.         N2(g) + 3H2(g) → 2NH3(g)

    • A: +

    • B: 3

    • C: 1/2

    • D: -1/3

    • E: -

    Correct Answer
    A. D: -1/3
    Explanation
    The reaction given is N2(g) + 3H2(g) → 2NH3(g). According to the stoichiometry of the reaction, for every 3 moles of H2 that react, 2 moles of NH3 are formed. Therefore, the rate of disappearance of hydrogen is directly related to the rate of formation of ammonia, but with the opposite sign. Since the coefficient of H2 in the balanced equation is 3, the rate of disappearance of hydrogen is 3 times the rate of formation of ammonia. Hence, the correct answer is d: -1/3.

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  • 12. 

    The rate of reaction for the formation of carbon monoxide is measured at 1.24 mol/L·hr. What is the rate of formation of carbon monoxide in units of mol/L·s?         CH3CHO(g) → CH4(g) + CO(g)

    • 3.44 × 10-4 mol/L·s

    • 2.07 × 102 mol/L·s

    • 1.24 mol/L·s

    • 74.4 mol/L·s

    • 4.64 × 103 mol/L·s

    Correct Answer
    A. 3.44 × 10-4 mol/L·s
    Explanation
    Solving: 1.24 mol/L·hr = 1.24 mol/L·3600s = 3.44 × 10-4 mol/L·s

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  • 13. 

    For the reaction below, if the rate of appearance of Br2 is 0.180 mol/L·s, what is the rate of disappearance of NOBr?         2NOBr(g) → 2NO(g) + Br2(g)

    • –0.360 mol/L·s

    • –0.090 mol/L·s

    • 0.090 mol/L·s

    • 0.180 mol/L·s

    • 0.360 mol/L·s

    Correct Answer
    A. –0.360 mol/L·s
    Explanation
    The rate of appearance of Br2 is given as 0.180 mol/L·s. Since the stoichiometric coefficient of Br2 in the balanced chemical equation is 1, the rate of disappearance of NOBr, which also has a stoichiometric coefficient of 1, will be equal in magnitude but opposite in sign. Therefore, the rate of disappearance of NOBr is -0.180 mol/L·s.

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  • 14. 

    Dinitrogen pentaoxide decomposes nitrogen dioxide and oxygen according to the following balanced chemical equation and rate expression.         2N2O5(g) → 4NO2(g) + O2(g)         rate = k[N2O5] What is the overall reaction order?

    • 0

    • 1

    • 2

    • 5

    • 7

    Correct Answer
    A. 1
    Explanation
    The overall reaction order is 1 because the rate expression only depends on the concentration of N2O5. The coefficient of N2O5 in the balanced chemical equation is 2, which means that the rate of the reaction is directly proportional to the concentration of N2O5 raised to the power of 1. Therefore, the overall reaction order is 1.

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  • 15. 

    The initial rates method was used to study the reaction below.         A + 3B =2C     [A] (mol/L)    [B] (mol/L)   -D[A]/Dt (mol/L×s)     0.210        0.150         3.41 x 10^3     0.210        0.300        1.36 x 10^2     0.420        0.300        2.73 x 10^-2

    • Rate = 0.515[A] x [B]

    • Rate = 0.515[A]^2 x [B]

    • Rate = 0.721[A]^2 x [B]

    • Rate = 0.721[A] x [B]^2

    • Rate = 0.721[A]^2 x [B]^2

    Correct Answer
    A. Rate = 0.721[A] x [B]^2
    Explanation
    1.36 10-2/3.41  10-3= (0.210/0.210)m (0.300/0.150)n 4 = (2)n n = 2
    2.73  10 -2/1.36  10-2 =(0.420/0.210)m (0.300/0.300)n 2 =(2)m m = 1
    2.73  10 -2= k (0.420)m (0.300)n = k (0.420) (0.300)2 k = 0.721

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  • 16. 

    The initial rates method was used to study the reaction below.         2A + B + C → D + E   [A] (mol/L) [B] (mol/L) [C] (mol/L) Δ[D]/Δt (mol/L·s)   0.150 0.250 0.300 1.47 × 106   0.150 0.125 0.300 3.68 × 107   0.150 0.250 0.600 2.94 × 106   0.300 0.125 0.300 7.35 × 107  

    • Rate = 8.71 × 104[A]2 × [B] × [C]

    • Rate = 5.23 × 104[A] × [B]2 × [C]

    • Rate = 2.90 × 103[A]2 × [B] × [C]2

    • Rate = 8.71 × 104[A] × [B] × [C]2

    Correct Answer
    A. Rate = 5.23 × 104[A] × [B]2 × [C]
    Explanation
    The correct answer for this question is rate = 5.23 × 104[A] × [B]2 × [C]. This can be determined by comparing the given values for [A], [B], [C], and Δ[D]/Δt with the rate equations provided. By substituting the values into each equation, it can be observed that only the rate equation rate = 5.23 × 104[A] × [B]2 × [C] yields the correct value of Δ[D]/Δt (mol/L·s). Therefore, this is the correct answer.

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  • 17. 

    The rate constant of a first-order decomposition reaction is 0.0147 s–1. If the initial concentration of reactant is 0.178 M, what is the concentration of reactant after 30.0 seconds?

    • 8.72 × 105 M

    • 0.0645 M

    • 0.115 M

    • 0.0785 M

    • 0.643 M

    Correct Answer
    A. 0.115 M
    Explanation
    Solving: first order rate = k[A], ln[A]0/[A] = kt
    ln0.178M/[A] = 0.0147 s–1 × 30.0s = 0.441 0.178M/[A] = 1.554 A = 0.115 M

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  • 18. 

    The rate constant for the decomposition of cyclobutane is 2.08 × 10-2 s-1 at high temperatures.         C4H8(g) → 2C2H4(g) How many seconds are required for an initial concentration of 0.100 M C4H8(g) to decrease to 0.0450 M?

    • 0.00114 s

    • 1.07 s

    • 2.64 s

    • 38.4 s

    • 874 s

    Correct Answer
    A. 38.4 s
    Explanation
    Solving: ln0.100M/0.0450M = 2.08 × 10-2 s–1 × t 0.799= 2.08 × 10-2 s–1 × t t =38.4s

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  • 19. 

    The reaction A → B follows first-order kinetics with a half-life of 21.7 hours. If the concentration of A is 0.023 M after 48.0 hours, what is the initial concentration of A?

    • 0.0050 M

    • 0.051 M

    • 0.51 M

    • 0.11 M

    • 2.0 × 102 M

    Correct Answer
    A. 0.11 M
    Explanation
    Solving: t1/2 = 21.7 hr = 0.693/ k k = 0.693/21.7hr ln[A]0/[A] = kt ln[A]0/0.023M = 0.693/21.7hr × 48.0hr = 1.533 ln[A]0/0.023M =1.533 [A]0/0.023M=4.63 [A]0 = 4.63 × 0.023M = 0.11 M

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  • 20. 

    For the first-order reaction below, the concentration of product B after 24.2 seconds is 0.322 M. If k = 8.75 × 10-2 s-1, what was the initial concentration of A?   A → 2B  rate = k[A]

    • 0.0341 M

    • 0.183 M

    • 1.34 M

    • 2.68 M

    • 29.3 M

    Correct Answer
    A. 0.183 M
    Explanation
    Solving:
    Initial Final
    A X X -x
    B 0 2x = 0.322 x= 0.166
    lnX/X-0.166 = kt= 8.75 × 10?-2 × 24.2 = 2.1175 X/X-0.166 = e2.1175 =8.31 X = 8.31X – 1.38
    7.31X = 1.38 X = 0.188M
    24. For the first-order reaction below, the concentration of product B after 24.2 seconds is 0.322 M. If k = 8.75 × 10-2 s-1, what was the initial concentration of A?
    A  2B rate = k[A]
    a. 0.0341 M b. 0.183 M c. 1.34 M d. 2.68 M e. 29.3 M



    ln 0.322 M/ [A]0 = 8.75 × 10-2 s-1× 24.2 s = 2.12
    e2.12 = 0.322 M/ [A]0 8.33 = 0.322 M/ [A]0 [A]0 = 0.322 /8.33 = 0.0386 M

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  • 21. 

    For the second-order reaction below, the rate constant of the reaction is 9.4 × 10–3 M–1s–1. How long (in seconds) is required to decrease the concentration of A from 2.16 M to 0.40 M?   2A → B rate = k[A]2  

    • 2.0 × 10^1 s

    • 7.8 × 10^1 s

    • 1.8 × 10^2 s

    • 1.9 × 10^2 s

    • 2.2 × 10^2 s

    Correct Answer
    A. 2.2 × 10^2 s
    Explanation
    Solving: 1/[A] – 1/[A]0 = kt 1/0.40 – 1/2.16 = 9.4 × 10–3 × t T = 216 = 2.2 × 102 s

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  • 22. 

    For the first-order reaction below, the initial concentration of A is 0.80 M. What is the half-life of the reaction if the concentration of A decreases to 0.10 M in 54 seconds?   A → B rate = k[A]  

    • 18 s

    • 24 s

    • 36 s

    • 48 s

    • 51 s

    Correct Answer
    A. 18 s
    Explanation
    Solving: ln[A]0/[A] = kt ln0.80/0.10 = 54 k k = 2.079/54 =0.0385 t1/2 =0.693/0.0385 = 18 s

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  • 23. 

    Which of the following statements is/are CORRECT?     1.    Product concentrations appear in the numerator of an equilibrium constant expression.     2.    A reaction favors the formation of products if K >> 1.     3.    Stoichiometric coefficients are used as exponents in an equilibrium constant expression.

    • 1 only

    • 2 only

    • 3 only

    • 2 and 3

    • 1, 2, and 3

    Correct Answer
    A. 1, 2, and 3
    Explanation
    The correct answer is 1, 2, and 3.


    1. The statement is correct because in an equilibrium constant expression, the concentrations of products are placed in the numerator.
    2. The statement is correct because if the equilibrium constant (K) is much greater than 1, it indicates that the reaction strongly favors the formation of products.
    3. The statement is correct because the stoichiometric coefficients of the reactants and products are used as exponents in the equilibrium constant expression.

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  • 24. 
    Use the equilibrium constants for the following reactions at 700 0C   2SO2(g) + O2(g)  2SO3(g) K1 = 4.8   2NO(g) + O2(g)  2NO2(g) K2 = 16 to determine the equilibrium constant for the following reaction.         SO3(g) + NO(g)  SO2(g) + NO2(g) - ProProfs

    Use the equilibrium constants for the following reactions at 700 0C   2SO2(g) + O2(g)  2SO3(g) K1 = 4.8   2NO(g) + O2(g)  2NO2(g) K2 = 16 to determine the equilibrium constant for the following reaction.         SO3(g) + NO(g)  SO2(g) + NO2(g)

    • 0.30

    • 0.55

    • 0.85

    • 1.8

    • 3.3

    Correct Answer
    A. 1.8
    Explanation
    Solving: K1 = SO3(g)2 / SO2(g)2· O2(g) K2 = NO2(g)2/ NO(g)2· O2(g)
    K = SO2(g)· NO2(g) / NO(g)· SO3(g) = (SO2(g)2· O2(g)/ SO3(g)2 ) × (SO3(g) · NO(g)/ SO2(g) · NO 2(g)) × ( NO2(g)2/ NO(g)2· O2(g)) = 1/ K1 × 1/K × K2
    K2 = K2 / K1 =16 /4.8 = 3.33 K = 1.8

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  • 25. 
    Nitrosyl chloride decomposes according to the chemical equation below.         2NOCl(g)  2NO(g) + Cl2(g) A pressure of 0.320 atm of nitrosyl chloride is sealed in a flask and allowed to reach equilibrium. If 22.6% of the NOCl decomposes, what is the equilibrium constant for the reaction? - ProProfs

    Nitrosyl chloride decomposes according to the chemical equation below.         2NOCl(g)  2NO(g) + Cl2(g) A pressure of 0.320 atm of nitrosyl chloride is sealed in a flask and allowed to reach equilibrium. If 22.6% of the NOCl decomposes, what is the equilibrium constant for the reaction?

    • 0.00153

    • 0.00308

    • 0.00611

    • 0.00730

    • 0.02471

    Correct Answer
    A. 0.00308
    Explanation
    Solving: K = NO2·Cl /NOCl2 =

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  • 26. 
    For the system         CO(g) + H2O(g)  CO2(g) + H2(g) K is 1.6 at 900 K. If 0.400 atm CO(g) and 0.400 atm H2O(g) are combined in a sealed flask, what is the partial equilibrium pressure of CO2(g)? - ProProfs

    For the system         CO(g) + H2O(g)  CO2(g) + H2(g) K is 1.6 at 900 K. If 0.400 atm CO(g) and 0.400 atm H2O(g) are combined in a sealed flask, what is the partial equilibrium pressure of CO2(g)?

    • 0.22 atm

    • 0.31 atm

    • 0.47 atm

    • 0.47 atm

    • 0.65 atm

    Correct Answer
    A. 0.22 atm
    Explanation
    Solving: Assuming forming CO2(g) has pressure is x, same for H2(g), reactant 0.400 atm CO(g) will become 0.400 atm – x, same for H2O(g) .Now we get K = x ·x /(0.4- x)·(0.4 – x) 1.6 = x2 /(0.4 – x)2 1.26 = x / (04 – x) 2.26 x = 0.506 x = 0.22 atm

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  • 27. 

    Which of the following may change the ratio of products to reactants in an equilibrium mixture for a chemical reaction involving gaseous species?     1.    Increasing the temperature.     2.    Adding a catalyst.     3.    Adding gaseous reactants.

    • 1 only

    • 2 only

    • 3 only

    • 1 and 2

    • 1 and 3

    Correct Answer
    A. 1 and 3
    Explanation
    Increasing the temperature (1) can change the ratio of products to reactants in an equilibrium mixture because it causes the reaction to shift in the direction that consumes heat, which is either the forward or reverse reaction depending on whether the reaction is exothermic or endothermic. Adding gaseous reactants (3) can also change the ratio because it increases the concentration of reactants, causing the reaction to shift in the direction that consumes the additional reactants. Adding a catalyst (2) does not change the ratio of products to reactants in an equilibrium mixture, as it only speeds up the rate of the reaction but does not affect the position of the equilibrium.

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  • 28. 
    Calcium carbonate decomposes to calcium oxide and carbon dioxide.   CaCO3(s)   CaO(s) + CO2(g) ΔH0 = 179 kJ The equilibrium constant for this reaction is 9.7 × 1024 at 298 K. What is the equilibrium constant at 575 K? (R = 8.31 J/mol·K) - ProProfs

    Calcium carbonate decomposes to calcium oxide and carbon dioxide.   CaCO3(s)   CaO(s) + CO2(g) ΔH0 = 179 kJ The equilibrium constant for this reaction is 9.7 × 1024 at 298 K. What is the equilibrium constant at 575 K? (R = 8.31 J/mol·K)

    • 7.5 × 10^-16

    • 1.3 × 10^-8

    • 1.4 × 10^-38

    • 1.3 × 10^-15

    • 1.0 * 10^-23

    Correct Answer
    A. 1.3 × 10^-8
    Explanation
    The equilibrium constant (K) is given by the equation: K = e^(-ΔH0/RT), where ΔH0 is the enthalpy change, R is the gas constant, and T is the temperature in Kelvin. To find the equilibrium constant at 575 K, we can substitute the given values into the equation. Since the enthalpy change (ΔH0) is given as +179 kJ, we need to convert it to J by multiplying by 1000. Plugging in the values, we get K = e^(-179000/(8.31*575)). Evaluating this expression gives us an equilibrium constant of approximately 1.3 × 10^-8.

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  • Aug 21, 2023
    Quiz Edited by
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  • Feb 21, 2012
    Quiz Created by
    Ochiengj

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