# 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

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? 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%

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

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