Grade 13 — Term 1
2 study notes · 11 practice questions · Aligned with NIE syllabus
Counted in the Idasara Academy app, which you open with a free account. The free notes, revision, practice and library pages for each lesson are linked below.
This term in the Idasara Academy app has 2 lessons (NIE Units 11, 12). For each lesson, this page gives you the points you must know and one exam-style question with its answer. None of it needs an account.
Lessons in This Term
Reaction Kinetics
What you must know
- Units of rate constant k: (mol dm⁻³)¹⁻ⁿ s⁻¹ (where n is the overall reaction order).
- Arrhenius equation: k = A e^(-Ea/RT) or ln(k) = ln(A) - Ea/(RT). The slope of a plot of ln(k) against 1/T is -Ea/R.
- Why does a 10 °C temperature rise roughly double a reaction rate?: A temperature rise of 10 °C roughly doubles the reaction rate because of the exponential increase in the fraction of molecules with E ≥ Ea.
Try one question
For the reaction 2A + B → C, initial rate experiments show that doubling [A] at constant [B] doubles the rate, while doubling [B] at constant [A] quadruples the rate. What is the overall reaction order and the derived SI units of the rate constant k?
- A. Overall order = 3, with rate constant units mol⁻² dm⁶ s⁻¹
- B. Overall order = 2, with rate constant units mol⁻¹ dm³ s⁻¹
- C. Overall order = 3, with rate constant units mol² dm⁻⁶ s⁻¹
- D. Overall order = 1, with rate constant units s⁻¹
Show the answer
Answer: A. Overall order = 3, with rate constant units mol⁻² dm⁶ s⁻¹. Rate ∝ [A]^m [B]^n. Doubling [A] doubles rate ⇒ 2^m = 2 ⇒ m = 1 (first order in A). Doubling [B] quadruples rate ⇒ 2^n = 4 ⇒ n = 2 (second order in B). Overall order = m + n = 1 + 2 = 3. Rate law: Rate = k [A] [B]². Units of k = (Rate) / ([A][B]²) = (mol dm⁻³ s⁻¹) / ((mol dm⁻³)(mol dm⁻³)²) = mol⁻² dm⁶ s⁻¹.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Chemical Equilibrium
What you must know
- What changes the numerical value of Kc and Kp?: Only temperature changes the numerical value of Kc and Kp; concentration and pressure shifts alter the reaction quotient Q without altering K.
- Exothermic reactions (ΔH < 0): increasing temperature shifts equilibrium backward, decreasing Kc.
- Endothermic reactions (ΔH > 0): increasing temperature shifts equilibrium forward, increasing Kc.
Try one question
At 700 K, the equilibrium constant K_c for the reaction H₂(g) + I₂(g) ⇌ 2HI(g) is 64.0. If 1.00 mol of H₂ and 1.00 mol of I₂ are introduced into an evacuated 1.00 dm³ rigid reaction vessel at 700 K, what is the equilibrium concentration of HI?
- A. 0.80 mol dm⁻³
- B. 0.40 mol dm⁻³
- C. 1.20 mol dm⁻³
- D. 1.60 mol dm⁻³
Show the answer
Answer: D. 1.60 mol dm⁻³. Let x mol dm⁻³ of H₂ react. At equilibrium: [H₂] = 1.00 - x, [I₂] = 1.00 - x, and [HI] = 2x. K_c = [HI]² / ([H₂][I₂]) = (2x)² / (1.00 - x)² = 64.0. Taking square root of both sides: 2x / (1.00 - x) = 8.00 ⇒ 2x = 8.00 - 8.00x ⇒ 10.00x = 8.00 ⇒ x = 0.80 mol dm⁻³. Therefore, [HI] = 2x = 2(0.80) = 1.60 mol dm⁻³.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Before you sign up
What does it cost?
I already go to a Chemistry tuition class
Then you are paying for Chemistry by the subject. Whether an Idasara plan would cost you less depends on how many classes you go to, so put in your own numbers.
Work it out on the pricing page →