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

Question

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

Answer/Explanation

Markscheme: B

Question

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

 

Answer/Explanation

Markscheme: D

Question

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%

▶️Answer/Explanation

Markscheme: C

The percentage error is calculated using the formula:

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

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

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

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

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

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

Question

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

▶️Answer/Explanation

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 \(10^\circ C\). 

Question

Which ionic compound has the most endothermic lattice enthalpy?

A.     NaCl

B.     KCl

C.     NaF

D.     KF

Answer/Explanation

Markscheme

C

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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