JEE Main 2014 · 11 Apr · Shift 1 · Chemistry Q33

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ChemistrySingle correct+4 / -1Question 33 of 90

In the reaction of formation of sulphur trioxide by contact process $\ce{2SO2 + O2 <=> 2SO3}$ the rate of reaction was measured as $\dfrac{d[\ce{O2}]}{dt}=-2.5\times10^{-4}$ mol L $^{-1}$ s $^{-1}$ . The rate of reaction in terms of $[\ce{SO2}]$ in mol L $^{-1}$ s $^{-1}$ will be :

-1.25\times10^{-4}

-2.50\times10^{-4}

-3.75\times10^{-4}

-5.00\times10^{-4}

Answer and worked solution

Official answer: D

$\ce{SO2}$ is used twice as fast as $\ce{O2}$ :
$\frac{d[\ce{SO2}]}{dt}=2\times(-2.5\times10^{-4})=-5.00\times10^{-4}\text{ mol L}^{-1}\text{s}^{-1}.$

Topics:Rate of reaction and rate law

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JEE Main 2014 · 11 Apr · Shift 1, Chemistry question 33