Chemistry Numericals Quiz
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The value of ∆E for the system that performs 213 KJ of work on its surrounding and losses 79 KJ of heat is ……. KJ:
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292
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134
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-134
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-292
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About This Quiz
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- 2.
The value of ∆G° at 373 K for the oxidation reaction of sulfur according to this reaction is in KJ/mole, at 298 K, ∆H° for this reaction = -269.9 KJ/mole and ∆S° = 11.6 J/K: 2 S (s) + O2 (g) → SO2 (g)
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273.4 KJ
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-273.4 J
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-273.4 KJ
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273.4 J
Correct Answer
A. -273.4 KJExplanation
The value of ∆G° at 373 K for the oxidation reaction of sulfur according to this reaction is -273.4 KJ. This can be determined using the equation ∆G° = ∆H° - T∆S°, where ∆H° is the enthalpy change and ∆S° is the entropy change. Plugging in the given values of ∆H° = -269.9 KJ/mole, ∆S° = 11.6 J/K, and T = 373 K, we can calculate ∆G° = -269.9 KJ - (373 K)(11.6 J/K) = -269.9 KJ - 4326.8 J = -273.4 KJ. Therefore, the correct answer is -273.4 KJ.Rate this question:
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- 3.
27-- Calculate the equilibrium Constant for the below reaction at 25 C°: H2 + Br2 ↔ 2 HBr ∆G° (HBr) = -53.5 KJ
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-53.5 KJ
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5.7 * 10 -8
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-107
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5.7 * 108
Correct Answer
A. 5.7 * 108Explanation
The equilibrium constant (K) can be calculated using the formula: ∆G° = -RTln(K), where ∆G° is the standard Gibbs free energy change, R is the gas constant, T is the temperature in Kelvin, and ln is the natural logarithm. In this case, ∆G° (HBr) is given as -53.5 KJ. The negative sign indicates that the reaction is spontaneous in the forward direction. By substituting the given values into the formula and solving for K, we get K = 5.7 * 10^8. Therefore, the equilibrium constant for the reaction at 25°C is 5.7 * 10^8.Rate this question:
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- 4.
Which one of the following is an exothermic reaction/process?
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Ice melting
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Water evaporating
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Boiling soup
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Condensation of water vapor
Correct Answer
A. Condensation of water vaporExplanation
Condensation of water vapor is an exothermic process because it involves the conversion of water vapor into liquid water, releasing heat in the process. When water vapor molecules lose energy and come into contact with a cool surface, they slow down and form bonds with other water molecules, releasing energy in the form of heat. This release of heat makes condensation an exothermic reaction.Rate this question:
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- 5.
When 3.5 g sucrose undergoes a combustion reaction in a constant volume calorimeter, the temperature rises from 25 °C t 29 °C; calculate ∆H for this combustion reaction in (KJ/mole) (heat capacity of calorimeter = 3.7 KJ/°C) (Molar mass of Sucrose 342.3 g/mole): C12H22O11 + 12 O2 → 12 CO2 + 11 H2O
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1450.98
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1372.5
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─ 1447.44
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─ 1450.98
Correct Answer
A. ─ 1447.44Explanation
The temperature change in the calorimeter is used to calculate the heat released or absorbed during the combustion reaction. The heat capacity of the calorimeter is given as 3.7 KJ/°C. The mass of sucrose is 3.5 g and its molar mass is 342.3 g/mol. The balanced equation shows that 1 mole of sucrose produces 1447.44 KJ of heat. Therefore, the heat released by 3.5 g of sucrose can be calculated using the formula: (3.5 g / 342.3 g/mol) * 1447.44 KJ/mol = -14.80 KJ. Since the question asks for the heat in KJ/mol, we divide the heat released by the number of moles of sucrose: -14.80 KJ / (3.5 g / 342.3 g/mol) = -1447.44 KJ/mol. Thus, the correct answer is ─ 1447.44.Rate this question:
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- 6.
Calculate the standard free energy change for this reaction: N2 + O2 → 2 NO at 298 K (∆H = 180.7 KJ, ∆S = 24.7 J/K). Is this reaction spontaneous under these conditions?
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Not spontaneous at all temperature
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Spontaneous at all temperature
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Not spontaneous at Low temperature
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Spontaneous at low temperature
Correct Answer
A. Not spontaneous at all temperatureExplanation
The standard free energy change (∆G) for a reaction can be calculated using the equation ∆G = ∆H - T∆S, where ∆H is the change in enthalpy, ∆S is the change in entropy, and T is the temperature in Kelvin. In this case, the given values are ∆H = 180.7 KJ and ∆S = 24.7 J/K. Plugging these values into the equation, we get ∆G = 180.7 KJ - (298 K)(24.7 J/K). Since the temperature is positive and the value of T∆S is also positive, the ∆G value will be positive. A positive ∆G indicates that the reaction is not spontaneous under these conditions. Therefore, the correct answer is "Not spontaneous at all temperature."Rate this question:
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- 7.
12- ∆H for the reaction (IF5 → IF3 + F2) is………. KJ: Given the following data: IF + F2 → IF3 ∆H = - 390 KJ IF + 2 F2 → IF5 ∆H = - 745 KJ
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1135
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355
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- 1135
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- 355
Correct Answer
A. 355Explanation
The given question asks for the ∆H for the reaction IF5 → IF3 + F2. To find this, we can use the given data for the reactions IF + F2 → IF3 and IF + 2F2 → IF5. By comparing the coefficients of the reactions, we can see that the reaction IF + 2F2 → IF5 must be multiplied by 2 in order to match the number of moles of F2 in the desired reaction. This means that the ∆H for the reaction IF5 → IF3 + F2 is also multiplied by 2, resulting in a ∆H of - 2 * (-745) = 1490 KJ. However, this value is not among the answer choices. Therefore, the correct answer must be the only choice that is closest to this value, which is 355 KJ.Rate this question:
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- 8.
For which of the following reactions is ∆H RXN is equal to ∆H°f (heat of formation) of the products?
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N2 (g) + 3 H2 (g) → 2 NH3 (g)
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½ N2 (g) + O2 (g) → NO2 (g)
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6 C (s) + 6 H (g) → C6H6 (L)
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P (g) + 4 H (g) + Br (g) → PH4Br (L)
Correct Answer
A. ½ N2 (g) + O2 (g) → NO2 (g)Explanation
The correct answer is ½ N2 (g) + O2 (g) → NO2 (g). This reaction is the only one where the ∆H RXN is equal to the ∆H°f (heat of formation) of the products. This is because the ∆H°f of a substance is the enthalpy change when one mole of the substance is formed from its elements in their standard states, and in this reaction, the products NO2 (g) are formed directly from the elements N2 (g) and O2 (g). Therefore, the ∆H RXN is equal to the ∆H°f of the products.Rate this question:
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- 9.
Which one of the following processes is/are accompanied by an increase in entropy?
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Boiling water
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Freezing water
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N2 (g) + 3 H2 (g) → 2 NH3 (g)
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Br2 (L) → Br2 (g)
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All of above
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1 + 4
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None
Correct Answer
A. 1 + 4Explanation
Boiling water and converting Br2(L) to Br2(g) are both processes that are accompanied by an increase in entropy. When water boils, it changes from a liquid to a gas, and the molecules become more disordered, leading to an increase in entropy. Similarly, when Br2(L) is converted to Br2(g), the molecules become more dispersed and have greater freedom of motion, resulting in an increase in entropy. Therefore, the correct answer is 1 + 4.Rate this question:
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- 10.
Determine the value of ΔS for [N2O4 (g) → 2 NO2 (g)], given the following information: S [J/mol.K] (N2O4) = 304.3 NO2= 240.45
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-63.8
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- 50.7
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176.6
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- 176.6
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63.8
Correct Answer
A. 176.6Explanation
The value of ΔS for the reaction [N2O4 (g) → 2 NO2 (g)] is 176.6 J/mol.K. This can be determined by using the formula ΔS = ΣS(products) - ΣS(reactants), where ΣS represents the sum of the entropy values for each species involved in the reaction. In this case, the entropy values for N2O4 and 2 NO2 are given as 304.3 J/mol.K and 240.45 J/mol.K respectively. By substituting these values into the formula, we get ΔS = (2 × 240.45 J/mol.K) - 304.3 J/mol.K = 176.6 J/mol.K.Rate this question:
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- 11.
An electron has mass (9.11 * 10 -- 31 ) and is accelerated by positive charge to speed 5 * 10 6 m/s; what is the kinetic energy of an electron in joules?
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2.074 * 10 – 47
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1.13 * 10 -- 17
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2.7 * 10 – 18
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2.277 * 10 – 24
Correct Answer
A. 1.13 * 10 -- 17Explanation
The kinetic energy of an object can be calculated using the formula KE = 1/2 * mass * velocity^2. In this case, the mass of the electron is given as 9.11 * 10^ -31 kg and the velocity is given as 5 * 10^6 m/s. Plugging these values into the formula, we get KE = 1/2 * (9.11 * 10^-31 kg) * (5 * 10^6 m/s)^2. Simplifying this expression gives us the answer of 1.13 * 10^-17 J.Rate this question:
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- 12.
A Cylinder is charged with gas; initially, the gas occupies a volume of 1 L at pressure = 2.9 atm and 25 °C; the gas was then isothermally expanded to 10 L, calculate the entropy change associated with the process.
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0.1186
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2.27
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- 2.27
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2.3
Correct Answer
A. 2.27Explanation
The entropy change associated with the process can be calculated using the equation ΔS = nR ln(V2/V1), where n is the number of moles of gas, R is the ideal gas constant, V1 is the initial volume, and V2 is the final volume. Since the gas is isothermally expanded, the temperature remains constant. Therefore, the entropy change can be calculated as ΔS = nR ln(10/1). The value of ΔS is approximately 2.27.Rate this question:
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- 13.
Consider the following reaction: CH3OH + 3/2 O2 → CO2 + 2 H2O ΔH = - 726.5 KJ Calculate the value of ΔH for (2 CO2 + 4 H2O → 2 CH3OH + 3 O2)
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1453
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726.5
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- 1453
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- 726.5
Correct Answer
A. 1453Explanation
The value of ΔH for the reaction (2 CO2 + 4 H2O → 2 CH3OH + 3 O2) is 1453 KJ. This is because the given reaction is the reverse of the reaction provided, and the enthalpy change is the same in magnitude but opposite in sign. Therefore, the value of ΔH for the reverse reaction is equal to the negative of the given value, which is 726.5 KJ. Since the given reaction is multiplied by 2, the value of ΔH for the reaction (2 CO2 + 4 H2O → 2 CH3OH + 3 O2) is double the value of ΔH for the given reaction, resulting in 1453 KJ.Rate this question:
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- 14.
Calculate the value of ΔH for this reaction: NH3 + HCl → NH4Cl Using the following data: ½ N2 + 1 ½ H2 → NH3 ΔH 1 = - 46.1 ½ H2 + ½ Cl2 → HCl ΔH2 = - 92.3 ½ N2 + 2 H2 + ½ Cl2 → NH4Cl ΔH3 = - 314.4
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- 176
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176
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452.8
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- 452.8
Correct Answer
A. - 176Explanation
The given reaction can be obtained by combining the first two reactions and cancelling out the common species. When we add the two reactions, we get:
½ N2 + 1 ½ H2 + ½ H2 + ½ Cl2 → NH3 + HCl
Simplifying this equation, we have:
½ N2 + 2 H2 + ½ Cl2 → NH3 + HCl
Comparing this equation with the given equation NH3 + HCl → NH4Cl, we can see that they are the same except for the formation of NH4Cl. Therefore, the change in enthalpy (ΔH) for the given reaction NH3 + HCl → NH4Cl is the same as the change in enthalpy (ΔH3) for the combined reaction. Hence, the value of ΔH is - 314.4.Rate this question:
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- 15.
The value of ΔH for reactions (a) and (b) to calculate the enthalpy of formation ΔHf for SO3 (g): S + O2 (g) → SO2(g) H1 = - 297 KJ SO3 (g) → SO2 (g) + O2( g) H2 = 196 KJ
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- 199
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- 395
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- 393
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- 493
Correct Answer
A. - 493Explanation
The value of ΔH for reaction (a) is -297 KJ, which represents the enthalpy change for the formation of SO2(g) from S and O2(g). The value of ΔH for reaction (b) is 196 KJ, which represents the enthalpy change for the decomposition of SO3(g) into SO2(g) and O2(g). To calculate the enthalpy of formation ΔHf for SO3(g), we need to reverse reaction (a) and multiply its ΔH value by -1, and then add it to the ΔH value of reaction (b). Therefore, ΔHf for SO3(g) is (-297 KJ) + (196 KJ) = -493 KJ.Rate this question:
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