OCR A-Level Chemistry A: Module 5 (Physical chemistry and transition elements) overview quiz quiz

Third

First

Second

Zero

$\text{rate} = k[A][B]$

$\text{rate} = k[A]$

$\text{rate} = k[A]^2$

$\text{rate} = k[B]$

Increasing the concentration of reactant

Adding more solvent

Increasing the temperature

Increasing the pressure of a solution reaction

$K_p = \frac{p(\text{N}_2)\,p(\text{H}_2)^3}{p(\text{NH}_3)^2}$

$K_p = \frac{p(\text{NH}_3)}{p(\text{N}_2)\,p(\text{H}_2)}$

$K_p = \frac{2\,p(\text{NH}_3)}{p(\text{N}_2) + 3\,p(\text{H}_2)}$

$K_p = \frac{p(\text{NH}_3)^2}{p(\text{N}_2)\,p(\text{H}_2)^3}$

1.30

7.00

12.70

13.70

The ethanoate ions react with the added $\text{H}^+$ to form ethanoic acid

The ethanoic acid molecules neutralise the added strong acid directly

The water present in the solution simply dilutes the added acid until it becomes weak

The sodium ions bond with and remove the added acid from solution

Smaller, because the heavier magnesium ions hold onto their charge more loosely than sodium ions do

Larger, because the ions are more highly charged and smaller, giving stronger attraction

The same, because both compounds are ionic and form regular giant lattices

Larger, only because oxygen is far more electronegative than chlorine in the lattice

50 K

100 K

5000 K

500 K

$+1.10\ \text{V}$

$+0.42\ \text{V}$

$-1.10\ \text{V}$

$-0.42\ \text{V}$

$[\text{Ar}]\,3d^4 4s^2$

$[\text{Ar}]\,3d^5 4s^1$

$[\text{Ar}]\,3d^6$

$[\text{Ar}]\,3d^4 4s^1$