IB DP Chemistry Reactivity 1.1 Measuring enthalpy changes IB Style Question Bank HL Paper 1


What is the enthalpy change for the following reaction?

A. -74 – 242 + 111
B. +74 + 242 + 111
C. -74 + 242 – 111
D. +74 + 484 – 222


Markscheme: B


Which diagram shows the enthalpy changes for dissolving a solid, MX, in water, if the process increases the temperature of the solution?



Markscheme: D


What is the percentage error if the enthalpy of combustion of a substance is determined experimentally to be -2100kJ \(mol^{-1}\) , but the literature value is -3500kJ \(mol^{-1}\)?

A. 80%
B. 60%
C. 40%
D. 20%


Markscheme: C

The percentage error is calculated using the formula:

\[ \text{Percentage Error} = \left| \frac{\text{Experimental Value} – \text{Literature Value}}{\text{Literature Value}} \right| \times 100\]

– Experimental Value = -2100 kJ/mol
– Literature Value = -3500 kJ/mol

\[ \text{Percentage Error} = \left| \frac{-2100 – (-3500)}{-3500} \right| \times 100\]

\[ \text{Percentage Error} = \left| \frac{1400}{3500} \right| \times 100\]

\[ \text{Percentage Error} = \frac{2}{5} \times 100\]

\[ \text{Percentage Error} = 40\%\]


When 100\(cm^3\) of 1.0mol \(dm^{-3}\) HCl is mixed with 100\(cm^3\) of 1.0mol \(dm^{-3}\) NaOH, the temperature of the resulting solution increases by 5.0°C. What will be the temperature change, in °C, when 50\(cm^3\) of 2.0mol \(dm^{-3}\) HCl is mixed with 50\(cm^{-3}\) of 2.0mol \(dm^{-3}\) NaOH?

A. 2.5
B. 5.0
C. 10
D. 20


Markscheme: C

Let’s reevaluate the solution:

The heat (\(q\)) evolved in a reaction can be related to the temperature change using the formula \(q = mc\Delta T\). Given that the reaction is \(HCl + NaOH \rightarrow NaCl + H_2O\) and assuming the specific heat capacity (\(c\)) of the resulting solution is approximately the same as that of water, we can calculate the heat evolved for the first reaction:

\[ q_1 = m_1c\Delta T_1 \]

For the second reaction, we have half the volume but double the concentration, so the moles of reacting substances remain the same. The heat evolved for the second reaction is:

\[ q_2 = m_2c\Delta T_2 \]

Since the moles (\(n\)) are the same for both reactions, and assuming the specific heat capacity (\(c\)) is the same, we can compare the temperature changes using the ratio:

\[ \frac{\Delta T_2}{\Delta T_1} = \frac{q_2}{q_1} \]

Given that \(\Delta T_1 = 5.0^\circ C\), we can solve for \(\Delta T_2\):

\[ \frac{\Delta T_2}{5.0} = \frac{q_2}{q_1} \]

Now, let’s reevaluate the given options:

A. 2.5: \(\frac{\Delta T_2}{5.0} = \frac{2.5}{5.0} = 0.5\)
B. 5.0: \(\frac{\Delta T_2}{5.0} = \frac{5.0}{5.0} = 1.0\)
C. 10: \(\frac{\Delta T_2}{5.0} = \frac{10}{5.0} = 2.0\)
D. 20: \(\frac{\Delta T_2}{5.0} = \frac{20}{5.0} = 4.0\)

The correct ratio is \(2.0\), and the temperature change for the second reaction (\(\Delta T_2\)) is \(10^\circ C\). 


Which ionic compound has the most endothermic lattice enthalpy?

A.     NaCl

B.     KCl

C.     NaF

D.     KF




Examiners report

This question on the effect of ionic radius on lattice enthalpy proved to be quite challenging, with a Difficulty Index of 43%, with many students considering that the larger potassium ion would give rise to a more endothermic lattice enthalpy than the smaller sodium ion.

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