IBDP Chemistry - Reactivity 1.2 Energy cycles in reactions- IB Style Questions For SL Paper 1A - FA 2025
▶️ Answer/Explanation
1. Calculate Oxidation State in \(NH_{3}\):
- Hydrogen is \(+1\) (with non-metals).
- \(N + 3(+1) = 0 \implies N = -3\).
2. Calculate Oxidation State in \(HNO_{2}\):
- Hydrogen is \(+1\). Oxygen is \(-2\).
- \(1(+1) + N + 2(-2) = 0\)
- \(1 + N – 4 = 0 \implies N – 3 = 0 \implies N = +3\).
3. Match to Options:
\(NH_3: -3\), \(HNO_2: +3\).
✅ Answer: (D)
▶️ Answer/Explanation
We use Hess’s law to build the target reaction from the given ones.
1. Reverse the first equation so that \(\text{Fe}_2\text{O}_3\) is a reactant and \(\text{Fe}\) is a product: \[ \text{Fe}_2\text{O}_3(s) \rightarrow 2\text{Fe}(s) + \tfrac{3}{2}\text{O}_2(g) \] Reversing changes the sign of the enthalpy: \[ \Delta H^{\circ}_1 = -x \]
2. Multiply the second equation by \(3\) so that we get \(3\text{CO}\) and \(3\text{CO}_2\): \[ 3\text{CO}(g) + \tfrac{3}{2}\text{O}_2(g) \rightarrow 3\text{CO}_2(g) \] The enthalpy change becomes: \[ \Delta H^{\circ}_2 = 3y \]
3. Add these two modified equations: \[ \begin{aligned} \text{Fe}_2\text{O}_3(s) &\rightarrow 2\text{Fe}(s) + \tfrac{3}{2}\text{O}_2(g) \\ 3\text{CO}(g) + \tfrac{3}{2}\text{O}_2(g) &\rightarrow 3\text{CO}_2(g) \end{aligned} \] Cancelling \(\tfrac{3}{2}\text{O}_2(g)\) on both sides gives: \[ \text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 3\text{CO}_2(g) + 2\text{Fe}(s) \] which is exactly the target reaction.
The overall enthalpy change is: \[ \Delta H^{\circ}_\text{reaction} = (-x) + 3y = 3y – x \]
✅ Answer: (A)
▶️ Answer/Explanation
Use the enthalpy of reaction formula based on formation enthalpies:
\[ \Delta H^\circ_{\text{rxn}} = \sum \Delta H_f^\circ(\text{products}) – \sum \Delta H_f^\circ(\text{reactants}) \]
Given:
\(\Delta H^\circ_{\text{rxn}} = -623\ \text{kJ}\) \(\Delta H_f^\circ(\text{H}_2\text{O}(l)) = -286\ \text{kJ}\) \(\Delta H_f^\circ(\text{N}_2(g)) = 0\) (elemental form) \(\Delta H_f^\circ(\text{O}_2(g)) = 0\)
Substitute into the formula: \[ -623 = \left[0 + 2(-286)\right] – \left[\Delta H_f^\circ(\text{N}_2\text{H}_4(l)) + 0\right] \]
Simplify: \[ -623 = -572 – \Delta H_f^\circ(\text{N}_2\text{H}_4(l)) \]
Solve for the unknown: \[ \Delta H_f^\circ(\text{N}_2\text{H}_4(l)) = -572 + 623 \]
✅ Answer: (C)