201. At constant pressure, a reaction that releases heat to the surroundings is described as
ⓐ. endothermic with \(\Delta H \gt 0\)
ⓑ. isochoric with \(\Delta H=0\)
ⓒ. adiabatic with \(\Delta H \gt 0\)
ⓓ. exothermic with \(\Delta H \lt 0\)
Correct Answer: exothermic with \(\Delta H \lt 0\)
Explanation: An exothermic reaction releases heat from the system to the surroundings. At constant pressure, the heat exchanged is equal to enthalpy change, \(q_p=\Delta H\). If the system releases heat, \(q_p\) is negative, so \(\Delta H\) is also negative. Endothermic changes have the opposite sign because the system absorbs heat. The word exothermic describes the direction of heat flow at constant pressure, not the speed of the reaction.
202. A process at constant pressure has \(\Delta H=+28\,\text{kJ mol}^{-1}\). The best interpretation is that the process is
ⓐ. exothermic because heat leaves the system
ⓑ. endothermic because heat is absorbed
ⓒ. isolated because \(\Delta H\) is positive
ⓓ. cyclic because enthalpy is a state function
Correct Answer: endothermic because heat is absorbed
Explanation: At constant pressure, \(\Delta H=q_p\) when only pressure-volume work is involved. A positive \(\Delta H\) means the system absorbs heat from the surroundings under this condition. Such a process is called endothermic. Exothermic processes have \(\Delta H \lt 0\) because heat is released by the system. A positive enthalpy change does not imply that the system is isolated. The sign tells the direction of heat transfer for the chosen system at constant pressure.
203. A reaction mixture in an open beaker becomes warm, and the surroundings gain heat. For the reaction mixture as the system, the enthalpy change at constant pressure is most likely
ⓐ. negative
ⓑ. positive
ⓒ. exactly zero
ⓓ. impossible to assign any sign
Correct Answer: negative
Explanation: If the surroundings gain heat, the reaction mixture has released heat. For the system, heat released is negative. In an open beaker at constant pressure, the heat change is connected with enthalpy change by \(q_p=\Delta H\). Therefore, \(\Delta H\) is negative for the reaction mixture. The warm beaker is an observation of heat transfer outward from the reacting system. The sign is assigned to the system, not to the surroundings.
204. Use the graph description below.
An energy profile is drawn with enthalpy on the vertical axis and progress of reaction on the horizontal axis. The products are at a lower enthalpy level than the reactants.
The reaction represented by the graph is
ⓐ. endothermic with \(\Delta H \gt 0\)
ⓑ. isothermal with \(\Delta H=0\)
ⓒ. impossible because products cannot have lower enthalpy
ⓓ. exothermic with \(\Delta H \lt 0\)
Correct Answer: exothermic with \(\Delta H \lt 0\)
Explanation: Enthalpy change for a reaction is given by the enthalpy of products minus the enthalpy of reactants. If products are lower in enthalpy than reactants, the difference is negative. A negative enthalpy change at constant pressure corresponds to heat release by the system. Such a reaction is exothermic. The graph does not mean the reaction has no energy change; it shows that energy is released as the system moves to lower enthalpy. Energy profile diagrams must be read using the relative levels of reactants and products.
205. For the change \(\mathrm{H_2O(s)}\rightarrow\mathrm{H_2O(l)}\) at its melting point, heat is absorbed by the system. The change is best classified as
ⓐ. exothermic, with \(\Delta H \lt 0\)
ⓑ. adiabatic, with \(q_p \gt 0\)
ⓒ. endothermic, \(\Delta H \gt 0\)
ⓓ. cyclic, with \(\Delta H=0\)
Correct Answer: endothermic, \(\Delta H \gt 0\)
Explanation: Melting requires heat absorption by the solid water system. At constant pressure, absorbed heat corresponds to a positive enthalpy change. Therefore, the melting of ice is an endothermic change with \(\Delta H \gt 0\). The chemical identity remains \(\mathrm{H_2O}\), but energy is still required to change the physical state. Exothermic changes release heat and would have \(\Delta H \lt 0\). A phase change can be thermodynamic even without a chemical reaction.
206. A reaction at constant pressure has \(q_p=-65\,\text{kJ}\) for the system. The reaction is
ⓐ. exothermic, and \(\Delta H=-65\,\text{kJ}\)
ⓑ. endothermic, and \(\Delta H=-65\,\text{kJ}\)
ⓒ. endothermic, and \(\Delta H=+65\,\text{kJ}\)
ⓓ. exothermic, and \(\Delta H=+65\,\text{kJ}\)
Correct Answer: exothermic, and \(\Delta H=-65\,\text{kJ}\)
Explanation: \( \textbf{Condition:} \) The process occurs at constant pressure.
\( \textbf{Constant-pressure relation:} \)
\[
q_p=\Delta H
\]
\( \textbf{Given heat:} \)
\[
q_p=-65\,\text{kJ}
\]
\( \textbf{Therefore:} \)
\[
\Delta H=-65\,\text{kJ}
\]
\( \textbf{Sign interpretation:} \) Negative heat for the system means heat is released.
\( \textbf{Classification:} \) A heat-releasing process at constant pressure is exothermic.
\( \textbf{Final answer:} \) The reaction is exothermic, and \(\Delta H=-65\,\text{kJ}\). The negative sign belongs to the system's heat change.
207. A student compares two constant-pressure reactions:
| Reaction | \(\Delta H\) |
|---|
| P | \(-120\,\text{kJ mol}^{-1}\) |
| Q | \(+45\,\text{kJ mol}^{-1}\) |
The correct comparison is
ⓐ. P is endothermic and Q is exothermic
ⓑ. both P and Q are exothermic
ⓒ. both P and Q are endothermic
ⓓ. P is exothermic and Q is endothermic
Correct Answer: P is exothermic and Q is endothermic
Explanation: At constant pressure, the sign of \(\Delta H\) shows whether heat is released or absorbed by the system. Reaction P has \(\Delta H \lt 0\), so it releases heat and is exothermic. Reaction Q has \(\Delta H \gt 0\), so it absorbs heat and is endothermic. The larger numerical magnitude of P does not change the sign rule. A negative sign must not be ignored when classifying a heat change. The classification follows the direction of heat flow, not just the size of the number.
208. Assertion: An endothermic reaction at constant pressure has \(\Delta H \gt 0\).
Reason: In an endothermic reaction, heat is absorbed by the system, and at constant pressure \(q_p=\Delta H\).
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: An endothermic reaction absorbs heat from the surroundings into the system. Heat absorbed by the system has a positive sign. At constant pressure, this heat is related to enthalpy change by \(q_p=\Delta H\). Therefore, an endothermic reaction at constant pressure has \(\Delta H \gt 0\). The Reason gives the sign convention and the constant-pressure relation needed to justify the Assertion. The conclusion would be incomplete if the system reference or constant-pressure condition were ignored.
209. A heat capacity value tells how much heat is required to raise the temperature of an object by
ⓐ. \(1\,\text{mol}\)
ⓑ. \(1\,\text{bar}\)
ⓒ. \(1\,\text{K}\)
ⓓ. \(1\,\text{L}\)
Correct Answer: \(1\,\text{K}\)
Explanation: Heat capacity \(C\) is the heat required to raise the temperature of a body or system by \(1\,\text{K}\). It is related to heat and temperature change by \(C=\frac{q}{\Delta T}\). Its unit is commonly \(\text{J K}^{-1}\). The value depends on the amount and nature of the substance for the body being considered. It is different from specific heat capacity, which is heat capacity per unit mass. The temperature interval \(1\,\text{K}\) has the same size as \(1\,{}^{\circ}\text{C}\) for temperature differences.
210. The relation for specific heat capacity is
ⓐ. \(c=\frac{q}{n\Delta T}\)
ⓑ. \(c=\frac{\Delta T}{q}\)
ⓒ. \(c=q\,m\,\Delta T\)
ⓓ. \(c=\frac{q}{m\Delta T}\)
Correct Answer: \(c=\frac{q}{m\Delta T}\)
Explanation: Specific heat capacity \(c\) is heat required per unit mass per unit temperature rise. The heat relation is \(q=mc\Delta T\), so \(c=\frac{q}{m\Delta T}\). Here \(m\) is mass and \(\Delta T\) is the temperature change. The unit is commonly \(\text{J g}^{-1}\text{K}^{-1}\) or \(\text{J kg}^{-1}\text{K}^{-1}\), depending on the mass unit used. The expression involving \(n\) gives molar heat capacity, not specific heat capacity. Keeping the basis as mass or moles prevents unit confusion.
211. A \(50.0\,\text{g}\) metal sample absorbs \(450\,\text{J}\) of heat, and its temperature rises by \(15.0\,\text{K}\). The specific heat capacity of the metal is
ⓐ. \(0.30\,\text{J g}^{-1}\text{K}^{-1}\)
ⓑ. \(0.60\,\text{J g}^{-1}\text{K}^{-1}\)
ⓒ. \(1.50\,\text{J g}^{-1}\text{K}^{-1}\)
ⓓ. \(6.00\,\text{J g}^{-1}\text{K}^{-1}\)
Correct Answer: \(0.60\,\text{J g}^{-1}\text{K}^{-1}\)
Explanation: \( \textbf{Known data:} \) \(q=450\,\text{J}\), \(m=50.0\,\text{g}\), \(\Delta T=15.0\,\text{K}\).
\( \textbf{Required quantity:} \) Specific heat capacity \(c\).
\( \textbf{Useful relation:} \)
\[
q=mc\Delta T
\]
\( \textbf{Rearrange:} \)
\[
c=\frac{q}{m\Delta T}
\]
\( \textbf{Substitute:} \)
\[
c=\frac{450\,\text{J}}{(50.0\,\text{g})(15.0\,\text{K})}
\]
\( \textbf{Denominator:} \)
\[
50.0\times15.0=750
\]
\( \textbf{Calculate:} \)
\[
c=\frac{450}{750}=0.60\,\text{J g}^{-1}\text{K}^{-1}
\]
\( \textbf{Final answer:} \) \(c=0.60\,\text{J g}^{-1}\text{K}^{-1}\). The mass and temperature rise both reduce the heat required per gram per kelvin.
212. A body has heat capacity \(C=120\,\text{J K}^{-1}\). If its temperature is raised from \(300\,\text{K}\) to \(305\,\text{K}\), the heat absorbed is
ⓐ. \(600\,\text{J}\)
ⓑ. \(24\,\text{J}\)
ⓒ. \(120\,\text{J}\)
ⓓ. \(36600\,\text{J}\)
Correct Answer: \(600\,\text{J}\)
Explanation: \( \textbf{Given heat capacity:} \) \(C=120\,\text{J K}^{-1}\).
\( \textbf{Temperature change:} \)
\[
\Delta T=305\,\text{K}-300\,\text{K}=5\,\text{K}
\]
\( \textbf{Heat relation:} \)
\[
q=C\Delta T
\]
\( \textbf{Substitution:} \)
\[
q=(120\,\text{J K}^{-1})(5\,\text{K})
\]
\( \textbf{Unit cancellation:} \) \(\text{K}^{-1}\times\text{K}\) cancels, leaving \(\text{J}\).
\( \textbf{Calculation:} \)
\[
q=600\,\text{J}
\]
\( \textbf{Final answer:} \) \(q=600\,\text{J}\). Heat capacity applies to the whole body, not per gram or per mole unless stated.
213. The unit \(\text{J mol}^{-1}\text{K}^{-1}\) belongs most directly to
ⓐ. heat capacity of a whole object
ⓑ. molar heat capacity
ⓒ. specific heat capacity
ⓓ. pressure-volume work only
Correct Answer: molar heat capacity
Explanation: Molar heat capacity is heat required to raise the temperature of \(1\,\text{mol}\) of a substance by \(1\,\text{K}\). Its common unit is \(\text{J mol}^{-1}\text{K}^{-1}\). Heat capacity of a whole object has unit \(\text{J K}^{-1}\). Specific heat capacity uses mass in the denominator, such as \(\text{J g}^{-1}\text{K}^{-1}\). Pressure-volume work has energy units such as \(\text{J}\), not heat capacity units. The unit reveals whether the heat quantity is per object, per gram, or per mole.
214. A sample contains \(2.0\,\text{mol}\) of a substance with molar heat capacity \(C_m=30\,\text{J mol}^{-1}\text{K}^{-1}\). The heat needed to raise its temperature by \(10\,\text{K}\) is
ⓐ. \(60\,\text{J}\)
ⓑ. \(300\,\text{J}\)
ⓒ. \(600\,\text{J}\)
ⓓ. \(1200\,\text{J}\)
Correct Answer: \(600\,\text{J}\)
Explanation: \( \textbf{Known data:} \) \(n=2.0\,\text{mol}\), \(C_m=30\,\text{J mol}^{-1}\text{K}^{-1}\), \(\Delta T=10\,\text{K}\).
\( \textbf{Useful relation:} \)
\[
q=nC_m\Delta T
\]
\( \textbf{Reason for relation:} \) Molar heat capacity is heat per mole per kelvin.
\( \textbf{Substitution:} \)
\[
q=(2.0\,\text{mol})(30\,\text{J mol}^{-1}\text{K}^{-1})(10\,\text{K})
\]
\( \textbf{Unit cancellation:} \) \(\text{mol}\) and \(\text{mol}^{-1}\) cancel, and \(\text{K}\) and \(\text{K}^{-1}\) cancel.
\( \textbf{Calculation:} \)
\[
q=600\,\text{J}
\]
\( \textbf{Final answer:} \) \(q=600\,\text{J}\). Multiplying by the number of moles is necessary because \(C_m\) is a per-mole quantity.
215. A table compares three heat-related quantities.
| Quantity | Formula | Usual unit |
|---|
| P. Heat capacity | \(C=\frac{q}{\Delta T}\) | \(\text{J K}^{-1}\) |
| Q. Specific heat capacity | \(c=\frac{q}{m\Delta T}\) | \(\text{J g}^{-1}\text{K}^{-1}\) |
| R. Molar heat capacity | \(C_m=\frac{q}{n\Delta T}\) | \(\text{J mol}^{-1}\text{K}^{-1}\) |
The correctly matched entries are
ⓐ. P only
ⓑ. P and Q only
ⓒ. Q and R only
ⓓ. P, Q, and R
Correct Answer: P, Q, and R
Explanation: Heat capacity \(C\) refers to the whole body or sample and has unit \(\text{J K}^{-1}\). Specific heat capacity \(c\) is per unit mass, so its formula includes \(m\) and its unit includes \(\text{g}^{-1}\) or \(\text{kg}^{-1}\). Molar heat capacity \(C_m\) is per mole, so its formula includes \(n\) and its unit includes \(\text{mol}^{-1}\). All three rows match the correct basis and unit. The same heat \(q\) may be interpreted differently depending on whether the basis is the whole sample, mass, or moles. The denominator shows what the heat value is being divided by.
216. Two samples of the same metal are at the same initial temperature. Sample P has mass \(20\,\text{g}\), and Sample Q has mass \(40\,\text{g}\). If both are heated through the same \(\Delta T\), the heat required for Q is
ⓐ. half that for P
ⓑ. twice that for P
ⓒ. equal to that for P
ⓓ. four times that for P
Correct Answer: twice that for P
Explanation: For the same substance, the specific heat capacity \(c\) is the same under the same conditions. The heat required is \(q=mc\Delta T\). If \(\Delta T\) is the same and \(c\) is the same, heat is directly proportional to mass. Sample Q has twice the mass of Sample P. Therefore, Q requires twice as much heat for the same temperature rise. The comparison uses proportional reasoning rather than a full numerical substitution.
217. A \(100\,\text{g}\) water sample is heated from \(25\,{}^{\circ}\text{C}\) to \(35\,{}^{\circ}\text{C}\). Take \(c=4.2\,\text{J g}^{-1}\text{K}^{-1}\). The heat absorbed is
ⓐ. \(4200\,\text{J}\)
ⓑ. \(420\,\text{J}\)
ⓒ. \(2100\,\text{J}\)
ⓓ. \(14700\,\text{J}\)
Correct Answer: \(4200\,\text{J}\)
Explanation: \( \textbf{Known data:} \) \(m=100\,\text{g}\), \(c=4.2\,\text{J g}^{-1}\text{K}^{-1}\).
\( \textbf{Temperature change:} \)
\[
\Delta T=35\,{}^{\circ}\text{C}-25\,{}^{\circ}\text{C}=10\,{}^{\circ}\text{C}
\]
\( \textbf{Temperature interval note:} \) A change of \(10\,{}^{\circ}\text{C}\) is equal in size to \(10\,\text{K}\).
\( \textbf{Heat relation:} \)
\[
q=mc\Delta T
\]
\( \textbf{Substitution:} \)
\[
q=(100\,\text{g})(4.2\,\text{J g}^{-1}\text{K}^{-1})(10\,\text{K})
\]
\( \textbf{Calculation:} \)
\[
q=4200\,\text{J}
\]
\( \textbf{Final answer:} \) \(q=4200\,\text{J}\). The temperature change is used, not the final temperature alone.
218. A sample absorbs the same heat \(q\) in two trials. In Trial P, its temperature rise is \(5\,\text{K}\). In Trial Q, its temperature rise is \(10\,\text{K}\). The heat capacity in Trial P compared with Trial Q is
ⓐ. half as large
ⓑ. twice as large
ⓒ. four times as large
ⓓ. the same only if \(q=0\)
Correct Answer: twice as large
Explanation: Heat capacity is given by \(C=\frac{q}{\Delta T}\). The heat absorbed is the same in both trials. If \(\Delta T\) is smaller, the calculated heat capacity is larger. Trial P has \(\Delta T=5\,\text{K}\), while Trial Q has \(\Delta T=10\,\text{K}\). Therefore, \(C_P=\frac{q}{5}\) and \(C_Q=\frac{q}{10}\), so \(C_P=2C_Q\). This comparison follows from the inverse relation between \(C\) and \(\Delta T\) for the same heat input.
219. A \(25.0\,\text{g}\) solid with \(c=0.80\,\text{J g}^{-1}\text{K}^{-1}\) absorbs \(400\,\text{J}\) of heat. Its temperature rise is
ⓐ. \(20\,\text{K}\)
ⓑ. \(10\,\text{K}\)
ⓒ. \(25\,\text{K}\)
ⓓ. \(40\,\text{K}\)
Correct Answer: \(20\,\text{K}\)
Explanation: \( \textbf{Given data:} \) \(m=25.0\,\text{g}\), \(c=0.80\,\text{J g}^{-1}\text{K}^{-1}\), \(q=400\,\text{J}\).
\( \textbf{Required:} \) Temperature rise \(\Delta T\).
\( \textbf{Heat relation:} \)
\[
q=mc\Delta T
\]
\( \textbf{Rearrange:} \)
\[
\Delta T=\frac{q}{mc}
\]
\( \textbf{Substitution:} \)
\[
\Delta T=\frac{400\,\text{J}}{(25.0\,\text{g})(0.80\,\text{J g}^{-1}\text{K}^{-1})}
\]
\( \textbf{Denominator:} \)
\[
25.0\times0.80=20.0\,\text{J K}^{-1}
\]
\( \textbf{Calculate:} \)
\[
\Delta T=\frac{400}{20.0}=20\,\text{K}
\]
\( \textbf{Final answer:} \) \(\Delta T=20\,\text{K}\). The units reduce to kelvin because heat is divided by heat capacity of the sample.
220. A claim says, “Specific heat capacity and molar heat capacity have the same unit because both measure heat needed for temperature rise.” The best correction is that
ⓐ. both must have unit \(\text{J K}^{-1}\) only
ⓑ. specific heat capacity is measured in \(\text{mol}\), while molar heat capacity is measured in \(\text{g}\)
ⓒ. specific heat capacity is per unit mass, while molar heat capacity is per mole
ⓓ. neither quantity contains a temperature unit
Correct Answer: specific heat capacity is per unit mass, while molar heat capacity is per mole
Explanation: Both quantities relate heat to temperature change, but their bases are different. Specific heat capacity is heat required per unit mass per unit temperature change, so units such as \(\text{J g}^{-1}\text{K}^{-1}\) are used. Molar heat capacity is heat required per mole per unit temperature change, so its unit is \(\text{J mol}^{-1}\text{K}^{-1}\). Heat capacity of an entire object is different again and has unit \(\text{J K}^{-1}\). The denominator identifies the physical basis of the quantity. Treating all heat-capacity units as identical hides whether the calculation should use mass or moles.
