# Difficult Chemistry Questions And Answers

Approved & Edited by ProProfs Editorial Team
At ProProfs Quizzes, our dedicated in-house team of experts takes pride in their work. With a sharp eye for detail, they meticulously review each quiz. This ensures that every quiz, taken by over 100 million users, meets our standards of accuracy, clarity, and engagement.
| Written by Ochiengj
O
Ochiengj
Community Contributor
Quizzes Created: 1 | Total Attempts: 1,327
Questions: 28 | Attempts: 1,338  Settings  Do you know everything about Chemistry? Take this difficult Chemistry quiz with questions and answers, and see how well you understand the subject. The difficulty level of this quiz is a bit high, so be prepared to answer. Go for this quiz and check out your Chemistry skills. This quiz will not only test your knowledge but help you enhance your knowledge. All the best! You can challenge your friends also by sharing your quiz scores. Go for it!

• 1.

### 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.

• A.

25.00C

• B.

31.50C

• C.

56.50C

• D.

32.30C

• E.

57.30C

C. 56.50C
Explanation
Solution: q = mass ×c ×Δt
2.71 kJ = 50.0 g× 1.72 J/g 0C × (Tf – 25 0C), Tf = 56.5 0C

Rate this question:

• 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?

• A.

0.866 J/g·0C

• B.

2.60 J/g·0C

• C.

0.650 J/g·0C

• D.

1.15 J/g·0C

• E.

2.83 × 105 J/g·0C

C. 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

Rate this question:

• 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?

• A.

32.00C

• B.

46.20C

• C.

47.20C

• D.

57.20C

• E.

65.0C

E. 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

Rate this question:

• 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.

• A.

0.055 J/g· 0C

• B.

0.29 J/g · 0C

• C.

0.69 J/g · 0C

• D.

0.18 J/g · 0C

• E.

2.6 J/g · 0C

B. 0.29 J/g · 0C
Explanation
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

Rate this question:

• 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.

• A.

55.37 C

• B.

58.60 C

• C.

59.40 C

• D.

60.80 C

• E.

64.19 C

B. 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

Rate this question:

• 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.

• A.

–116 J/g

• B.

–934 J/g

• C.

–1.03 × 103 J/g

• D.

–2.19 × 103 J/g

• E.

–1.03 × 104 J/g

C. –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

Rate this question:

• 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.

• A.

0.00 0C

• B.

2.24 0C

• C.

5.22 0C

• D.

25.0 0C

• E.

47.8 0C

B. 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

Rate this question:

• 8.

### All of the following statements are true EXCEPT

• A.

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

• B.

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

• C.

A reaction with a negative enthalpy is exothermic.

• D.

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

• E.

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

E. 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.

Rate this question:

• 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

• A.

+30.24 kJ

• B.

–206.9 kJ

• C.

–298.6 kJ

• D.

–1398.8 kJ

• E.

–1663.6 kJ

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

Rate this question:

• 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)?

• A.

–10.3 kJ

• B.

–20.7 kJ

• C.

–29.8 kJ

• D.

–43.7 kJ

• E.

–65.6 kJ

C. –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

Rate this question:

• 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.

A: +

• B.

B: 3

• C.

C: 1/2

• D.

D: -1/3

• E.

E: -

D. 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.

Rate this question:

• 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)

• A.

3.44 × 10-4 mol/L·s

• B.

2.07 × 102 mol/L·s

• C.

1.24 mol/L·s

• D.

74.4 mol/L·s

• E.

4.64 × 103 mol/L·s

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

Rate this question:

• 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)

• A.

–0.360 mol/L·s

• B.

–0.090 mol/L·s

• C.

0.090 mol/L·s

• D.

0.180 mol/L·s

• E.

0.360 mol/L·s

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.

Rate this question:

• 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?

• A.

0

• B.

1

• C.

2

• D.

5

• E.

7

B. 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.

Rate this question:

• 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

• A.

Rate = 0.515[A] x [B]

• B.

Rate = 0.515[A]^2 x [B]

• C.

Rate = 0.721[A]^2 x [B]

• D.

Rate = 0.721[A] x [B]^2

• E.

Rate = 0.721[A]^2 x [B]^2

D. 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

Rate this question:

• 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

• A.

Rate = 8.71 × 104[A]2 × [B] × [C]

• B.

Rate = 5.23 × 104[A] × [B]2 × [C]

• C.

Rate = 2.90 × 103[A]2 × [B] × [C]2

• D.

Rate = 8.71 × 104[A] × [B] × [C]2

B. 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.

Rate this question:

• 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?

• A.

8.72 × 105 M

• B.

0.0645 M

• C.

0.115 M

• D.

0.0785 M

• E.

0.643 M

C. 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

Rate this question:

• 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?

• A.

0.00114 s

• B.

1.07 s

• C.

2.64 s

• D.

38.4 s

• E.

874 s

D. 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

Rate this question:

• 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?

• A.

0.0050 M

• B.

0.051 M

• C.

0.51 M

• D.

0.11 M

• E.

2.0 × 102 M

D. 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

Rate this question:

• 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]

• A.

0.0341 M

• B.

0.183 M

• C.

1.34 M

• D.

2.68 M

• E.

29.3 M

B. 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

Rate this question:

• 21.

### For the second-order reaction below, the rate constant of the reaction is 9.4 × 10–3M–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

• A.

2.0 × 10^1 s

• B.

7.8 × 10^1 s

• C.

1.8 × 10^2 s

• D.

1.9 × 10^2 s

• E.

2.2 × 10^2 s

E. 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

Rate this question:

• 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]

• A.

18 s

• B.

24 s

• C.

36 s

• D.

48 s

• E.

51 s

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

Rate this question:

• 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.

• A.

1 only

• B.

2 only

• C.

3 only

• D.

2 and 3

• E.

1, 2, and 3

E. 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.

Rate this question:

• 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)

• A.

0.30

• B.

0.55

• C.

0.85

• D.

1.8

• E.

3.3

D. 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

Rate this question:

• 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?

• A.

0.00153

• B.

0.00308

• C.

0.00611

• D.

0.00730

• E.

0.02471

B. 0.00308
Explanation
Solving: K = NO2·Cl /NOCl2 =

Rate this question:

• 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)?

• A.

0.22 atm

• B.

0.31 atm

• C.

0.47 atm

• D.

0.47 atm

• E.

0.65 atm

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

Rate this question:

• 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.

• A.

1 only

• B.

2 only

• C.

3 only

• D.

1 and 2

• E.

1 and 3

E. 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.

Rate this question:

• 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)

• A.

7.5 × 10^-16

• B.

1.3 × 10^-8

• C.

1.4 × 10^-38

• D.

1.3 × 10^-15

• E.

1.0 * 10^-23 Back to top