In a reaction \(A\) and \(B\) reacts to form product. The initial rate of reaction \((r_0)\) was determined using different initial concentration of \(A\) and \(B\).
| \(A/\, \ mol L^{-1}\) | 0.10 | 0.10 | 0.20 |
| \(B/\, \ mol L^{-1}\) | 0.30 | 0.10 | 0.30 |
| \(r_0/\, \ mol L^{-1}\) | \(6.81 \times 10^{-4}\) | \(2.27 \times 10^{-4}\) | \(13.62 \times 10^{-4}\) |
What is the initial rate of reaction \((r_0)\) when critical concentration of \(A\) and \(B\) is \(0.50\, \ mol/L\) and \(0.50\, \ mol/L\) respectively.
Answer & explanation
Correct answer: option 1
The correct answer is option 1. \(56.75 \times 10^{-4}\, \ molL^{-1}s^{-1}\).
Let the rate law of the given reaction be
\(r = k[A]^x[B]^y\)
From the given data, we get
\(6.81 \times 10^{-4} = k[0.10]^x[0.30]^y\) -------(1)
\(2.27 \times 10^{-4} = k[0.10]^x[0.10]^y\) -------(2)
\(13.62 \times 10^{-4} = k[0.20]^x[0.30]^y\) -------(3)
Dividing equation (1) by equation (2), we get
\(\frac{6.81 \times 10^{-4}}{2.27 \times 10^{-4}} = \frac{k[0.10]^x[0.30]^y}{ k[0.10]^x[0.10]^y}\)
\(⇒ \frac{6.81}{2.27} = \left(\frac{0.30}{0.10}\right)^y\)
\(⇒ (3)^1 = (3)^y\)
\(⇒ y = 1\)
Dividing equation (3) by (1), we get
\(\frac{13.62 \times 10^{-4}}{6.81 \times 10^{-4}} = \frac{k[0.20]^x[0.30]^y}{ k[0.10]^x[0.30]^y}\)
\(⇒ \frac{13.62}{2.6.81} = \left(\frac{0.20}{0.10}\right)^x\)
\(⇒ (2)^1 = (2)^x\)
\(⇒ x = 1\)
Let us now put the values of \(x\) and \(y\) in equation (1) to get the value of the rate constant, \(k\)
\(6.81 \times 10^{-4} = k[0.10]^1[0.30]^1\)
\(⇒ 6.81 \times 10^{-4} = k \times 0.03\)
\(⇒ k = \frac{6.81 \times 10^{-4}}{0.03}\)
\(⇒ k = 227 \times 10^{-1}\)
When critical concentration of \(A\) and \(B\) is \(0.50\, \ mol/L\) and \(0.50\, \ mol/L\) respectively, then
\(r = 227 \times 10^{-1} [0.50]^1[0.50]1\)
\(⇒ r = 227 \times 10^{-1} \times 0.50 \times 0.50\)
\(⇒ r = 56.75 \times 10^{-4}\)