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Question: In the formation of sulphur trioxide by the contact process, $SO_2 + \frac{1}{2}O_2 \rightleftharpoo...

In the formation of sulphur trioxide by the contact process, SO2+12O2⇌SO3SO_2 + \frac{1}{2}O_2 \rightleftharpoons SO_3, the rate of reaction can be measured as −d[SO2]dt=6.0×10−4\frac{-d[SO_2]}{dt} = 6.0 \times 10^{-4} mol L−1^{-1}s−1^{-1} Here the correct statements are

A

The rate of disappearance of O2O_2 will be 4.0 ×10−4\times 10^{-4} mol L−1^{-1}s−1^{-1}

B

The rate of reaction will be 6.0 ×10−4\times 10^{-4} mol L−1^{-1}s−1^{-1}

C

The rate of reaction expressed in terms of SO3SO_3 will be 6.0 ×10−4\times 10^{-4} mol L−1^{-1}s−1^{-1}

D

The rate of disappearance of O2O_2 will be 3.0 ×10−4\times 10^{-4} mol L−1^{-1}s−1^{-1}

Answer

B, C, D

Explanation

Solution

The given reaction is:

SO2+12O2⇌SO3SO_2 + \frac{1}{2}O_2 \rightleftharpoons SO_3

The general expression for the rate of reaction for a balanced chemical equation aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD is:

Rate =−1ad[A]dt=−1bd[B]dt=+1cd[C]dt=+1dd[D]dt= -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} = +\frac{1}{d}\frac{d[D]}{dt}

Applying this to the given reaction:

Rate =−11d[SO2]dt=−11/2d[O2]dt=+11d[SO3]dt= -\frac{1}{1}\frac{d[SO_2]}{dt} = -\frac{1}{1/2}\frac{d[O_2]}{dt} = +\frac{1}{1}\frac{d[SO_3]}{dt}

Rate =−d[SO2]dt=−2d[O2]dt=+d[SO3]dt= -\frac{d[SO_2]}{dt} = -2\frac{d[O_2]}{dt} = +\frac{d[SO_3]}{dt}

We are given that the rate of disappearance of SO2SO_2 is −d[SO2]dt=6.0×10−4\frac{-d[SO_2]}{dt} = 6.0 \times 10^{-4} mol L−1^{-1}s−1^{-1}.

Now let's evaluate each statement:

(A) The rate of disappearance of O2O_2 will be 4.0 ×10−4\times 10^{-4} mol L−1^{-1}s−1^{-1}

From the rate expression, we have:

−d[SO2]dt=−2d[O2]dt-\frac{d[SO_2]}{dt} = -2\frac{d[O_2]}{dt}

Given −d[SO2]dt=6.0×10−4\frac{-d[SO_2]}{dt} = 6.0 \times 10^{-4} mol L−1^{-1}s−1^{-1}.

So, 6.0×10−4=2×(−d[O2]dt)6.0 \times 10^{-4} = 2 \times \left(-\frac{d[O_2]}{dt}\right)

The rate of disappearance of O2O_2 is −d[O2]dt-\frac{d[O_2]}{dt}.

−d[O2]dt=6.0×10−42=3.0×10−4-\frac{d[O_2]}{dt} = \frac{6.0 \times 10^{-4}}{2} = 3.0 \times 10^{-4} mol L−1^{-1}s−1^{-1}.

Therefore, statement (A) is incorrect.

(B) The rate of reaction will be 6.0 ×10−4\times 10^{-4} mol L−1^{-1}s−1^{-1}

By definition, the rate of reaction R=−d[SO2]dtR = -\frac{d[SO_2]}{dt}.

Given −d[SO2]dt=6.0×10−4-\frac{d[SO_2]}{dt} = 6.0 \times 10^{-4} mol L−1^{-1}s−1^{-1}.

So, the rate of reaction is 6.0×10−46.0 \times 10^{-4} mol L−1^{-1}s−1^{-1}.

Therefore, statement (B) is correct.

(C) The rate of reaction expressed in terms of SO3SO_3 will be 6.0 ×10−4\times 10^{-4} mol L−1^{-1}s−1^{-1}

From the rate expression, the rate of reaction R=+d[SO3]dtR = +\frac{d[SO_3]}{dt}.

Since R=6.0×10−4R = 6.0 \times 10^{-4} mol L−1^{-1}s−1^{-1} (from statement B), then d[SO3]dt=6.0×10−4\frac{d[SO_3]}{dt} = 6.0 \times 10^{-4} mol L−1^{-1}s−1^{-1}.

This means the rate of formation of SO3SO_3 is 6.0×10−46.0 \times 10^{-4} mol L−1^{-1}s−1^{-1}. The statement refers to the rate of reaction expressed in terms of SO3SO_3, which implies the rate of formation of SO3SO_3.

Therefore, statement (C) is correct.

(D) The rate of disappearance of O2O_2 will be 3.0 ×10−4\times 10^{-4} mol L−1^{-1}s−1^{-1}

As calculated in the evaluation of statement (A), the rate of disappearance of O2O_2 (−d[O2]dt-\frac{d[O_2]}{dt}) is 3.0×10−43.0 \times 10^{-4} mol L−1^{-1}s−1^{-1}.

Therefore, statement (D) is correct.

The correct statements are (B), (C), and (D).