Question

The following results were obtained during study of the reaction \(2NO(g) + Cl_2(g) \rightarrow 2NOCl(g)\). Determine the value of \([X]\) in mol/L.

• A. \([X] = 0.8\)
• B. \([X] = 0.3\)
• C. \([X] = 0.4\)
• D. \([X] = 0.5\)

Hint:

In rate law problems, compare experiments where only one reactant concentration changes. This helps determine the order with respect to that reactant easily.

Answer 1

The Correct Option is D

Step 1: Write the general rate law.
For the reaction:
\[ 2NO(g) + Cl_2(g) \rightarrow 2NOCl(g) \] Assume the rate law as:
\[ \text{Rate} = k[NO]^m[Cl_2]^n \] where \(m\) is the order with respect to \(NO\), and \(n\) is the order with respect to \(Cl_2\).

Step 2: Find order with respect to \(Cl_2\).
Compare experiments I and II.
In both experiments, concentration of \(NO\) remains same:
\[ [NO] = 0.2 \] But concentration of \(Cl_2\) changes from \(0.2\) to \(0.4\), meaning it is doubled.
Rate changes from:
\[ 6.0 \times 10^{-3} \rightarrow 2.4 \times 10^{-2} \] \[ \frac{2.4 \times 10^{-2}}{6.0 \times 10^{-3}} = 4 \] So, when \([Cl_2]\) is doubled, rate becomes 4 times.
\[ 2^n = 4 \] \[ n = 2 \]

Step 3: Find order with respect to \(NO\).
Compare experiments I and III.
In both experiments, concentration of \(Cl_2\) remains same:
\[ [Cl_2] = 0.2 \] But concentration of \(NO\) changes from \(0.2\) to \(0.4\), meaning it is doubled.
Rate changes from:
\[ 6.0 \times 10^{-3} \rightarrow 1.2 \times 10^{-2} \] \[ \frac{1.2 \times 10^{-2}}{6.0 \times 10^{-3}} = 2 \] So, when \([NO]\) is doubled, rate becomes 2 times.
\[ 2^m = 2 \] \[ m = 1 \]

Step 4: Write the final rate law.
Therefore, the rate law is:
\[ \text{Rate} = k[NO][Cl_2]^2 \]

Step 5: Calculate rate constant using experiment I.
From experiment I:
\[ [NO] = 0.2, \quad [Cl_2] = 0.2, \quad \text{Rate} = 6.0 \times 10^{-3} \] Substitute in rate law:
\[ 6.0 \times 10^{-3} = k(0.2)(0.2)^2 \] \[ 6.0 \times 10^{-3} = k(0.2)(0.04) \] \[ 6.0 \times 10^{-3} = k(0.008) \] \[ k = \frac{6.0 \times 10^{-3}}{0.008} \] \[ k = 0.75 \]

Step 6: Use experiment IV to calculate \(X\).
From experiment IV:
\[ [NO] = X, \quad [Cl_2] = 0.6, \quad \text{Rate} = 1.35 \times 10^{-1} \] Using the rate law:
\[ 1.35 \times 10^{-1} = 0.75(X)(0.6)^2 \] \[ 0.135 = 0.75(X)(0.36) \] \[ 0.135 = 0.27X \]

Step 7: Solve for \(X\).
\[ X = \frac{0.135}{0.27} \] \[ X = 0.5 \] Final Answer:
\[ \boxed{[X] = 0.5 \, \text{mol/L}} \] Hence, the correct answer is option (D).