The first order rate constant for the decomposition of CaCO₃ at 700 K is 6.36 × 10⁻³s⁻¹ and activation energy is 209 kJ mol⁻¹. Its rate constant (in s⁻¹) at 600 K is x × 10⁻⁶. The value of x is ___________. (Nearest integer)
[Given R = 8.31 J K⁻ mol⁻¹; log 6.36 × 10⁻³ = -2.19, 10 -4.79 = 1.62 × 10⁻⁵]

Solution

Chemistry ma’am · 4 steps

Use the log form of Arrhenius, add the drop in log k to -2.19, and the given antilog finishes it.

1. Two-temperature form. Write the Arrhenius equation at 700 K and at 600 K and take the ratio, so A cancels. Here T₁ = 700 K with k₁ = 6.36 × 10⁻³ s⁻¹, and T₂ = 600 K.

   log k₂/k₁ = Eₐ/2.303R(1/T₁-1/T₂)

2. Put in numbers. The bracket is 1/700-1/600 = -100/420000 = -2.381 × 10⁻⁴ K⁻¹. Also 2.303 × 8.31 = 19.14. So log k₂/k₁ = (209000 × (-2.381 × 10⁻⁴))/19.14 = -2.60.
3. Find log k2. log k₂ = log k₁ - 2.60 = -2.19 - 2.60 = -4.79. The given antilog gives k₂ = 10^(-4.79) = 1.62 × 10⁻⁵ s⁻¹.
4. So. k₂ = 16.2 × 10⁻⁶ s⁻¹, so x rounds to 16.

The common slip: 2. Reading x straight off 1.62 × 10⁻⁵ gives 2. The question wants the answer as x × 10⁻⁶, so rewrite it as 16.2 × 10⁻⁶ first.

The idea to keep. Cooling a reaction makes k smaller, so log k must drop. Keep Eₐ in joules to match R, and fit the final k to the power of ten the question asks for.

Answer from the NTA final answer key. Solution by PrepWiser.