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JEE Advanced 2019Paper 1
Exam
JEE Advanced
Year
2019
Paper
Paper 1
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Advanced 2019 · Paper 1 · Chemistry · Chemical Kinetics

Consider the kinetic data given in the following table for the...

Consider the kinetic data given in the following table for the reaction A + B + C \toProduct.<br/><tableborder="1"><tbody><tr><td>ExperimentNo.</td><td>[A](moldm−3)</td><td>[B](moldm−3)Product.<br /> <table border="1"> <tbody> <tr> <td>Experiment No.</td> <td>\left[A\right]\left(mol{dm}^{- 3}\right)</td> <td>\left[B\right] \left(mold{m}^{- 3}\right) \left[C\right] \left(mold{m}^{- 3}\right)</td><td>Rateofreaction(moldm−3s−1)</td> <td>Rate of reaction \left(mold{m}^{- 3}{s}^{- 1}\right) 11 0.20.2 0.10.1 0.10.1 6.0×10−56.0 \times {10}^{- 5} 22 0.20.2 0.20.2 0.10.1 6.0×10−56.0 \times {10}^{- 5} 33 0.20.2 0.10.1 0.20.2 1.2×10−41.2 \times {10}^{- 4} 44 0.30.3 0.10.1 0.10.1 9.0×10−59.0 \times {10}^{- 5}
The rate of the reaction for \left[A\right] = 0.15mold{m}^{- 3}, \left[B\right] = 0.25mold{m}^{- 3} and \left[C\right] = 0.15mold{m}^{- 3} is found to be Y×10−5moldm−3s−1.Y \times {10}^{- 5}mold{m}^{- 3}{s}^{- 1} . The value of YY is __________

Correct answer

0

Step-by-step solution

r=k[A]a[B]b[C]c\text{r} = \text{k} [ \text{A} ] {}^{\text{a}}{[ \text{B} ]}^{\text{b}}{[ \text{C} ]}^{\text{c}}
6×10−5=k(0.2)a(0.1)b(0.1)c6 \times {10}^{- 5} = \text{k}{( 0.2 )}^{\text{a}}{( 0.1 )}^{\text{b}}{( 0.1 )}^{\text{c}} ..... (1)
6×10−5=k(0.2)a(0.2)b(0.1)c6 \times {10}^{- 5} = k{( 0.2 )}^{\text{a}}{( 0.2 )}^{\text{b}}{( 0.1 )}^{\text{c}} ...... (2)
1.2×10−4=k(0.2)a(0.1)b(0.2)c1.2 \times {10}^{- 4} = k{( 0.2 )}^{\text{a}}{( 0.1 )}^{\text{b}}{( 0.2 )}^{\text{c}} ...... (3)
9×10−5=k(0.3)a(0.1)b(0.1)c9 \times {10}^{- 5} = k{( 0.3 )}^{\text{a}}{( 0.1 )}^{\text{b}}{( 0.1 )}^{\text{c}} ...... (4)
From equation (1) and (ii) : b=0b = 0
From equation (1) and (iii) : c=1c = 1
From equation (1) and (iv) : a=1a = 1
Put x,yx, y and zz in equation (i) we get;
6×10−5=k(0.2)1(0.1)0(0.1)16 \times {10}^{- 5} = k{( 0.2 )}^{1}{( 0.1 )}^{0}{( 0.1 )}^{1}
r=3×10−3[A][C]\text{r} = 3 \times {10}^{- 3} [ \text{A} ] [ \text{C} ]
For [A]=0.15moldm−3[ \text{A} ] = 0.15\text{mol}{\text{dm}}^{- 3}
[B]=0.25moldm−3[ \text{B} ] = 0.25\text{mol}{\text{dm}}^{- 3}
[C]=0.15moldm−3[ \text{C} ] = 0.15\text{mol}{\text{dm}}^{- 3}
r=3×10−3×0.15×0.15\text{r} = 3 \times {10}^{- 3} \times 0.15 \times 0.15
r=6.75×10−5\text{r} = 6.75 \times {10}^{- 5}
∴r=6.75∴ \text{r}\text{=}\text{6.75}
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