Question

In the formation of sulphur trioxide by the contact process, 2SO₂ + O₂2SO₃, the rate of reaction was measured as

–$\frac{d\lbrack O_{2}\rbrack}{dt}$=3.0 × 10^–4mol L^–1s^–1.

The rate of reaction expressed in terms of SO₃ will be -

• A. 3.0 × 10^–4mol L^–1s^–1
• B. 6.0 × 10^–4mol L^–1s^–1
• C. 1.5 × 10^–4mol L^–1s^–1
• D. 4.5 × 10^–4mol L^–1s^–1

Answer 1

Answer: (B)

6.0 × 10^–4mol L^–1s^–1

Answer 2

$–\frac{d\lbrack O_{2}\rbrack}{dt}$ = $+ \frac{1}{2}$ $\frac{d\lbrack SO_{3}\rbrack}{dt}$

3 × 10^–4 = $\frac{1}{2}$ $\frac{d\lbrack SO_{3}\rbrack}{dt}$

$\frac{d\lbrack SO_{3}\rbrack}{dt}$ = 6 × 10^–4 M/sec