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  • for a hypothetical reaction <img alt="A+3B\rightarrow P\: \: \: \: \: \: \: \Delta H=-2x\: \: kJ/mole\: of\: A" src="https://entrancecorner.oncodecogs.com/gif.latex?A&plus;3B%5Crighta

for a hypothetical reaction

A+3B\rightarrow P\: \: \: \: \: \: \: \Delta H=-2x\: \: kJ/mole\: of\: A

and   M\rightarrow 2Q+R\: \: \: \: \: \: \: \Delta H=+x\: \: kJ/mole\: of\: M

If these reactions are carried simultaneously in a reactor such that temperature is not changing. If rate of disappearance of B is y\: M\: sec^{-1} then rate of formation (in\: M\: sec^{-1}) of Q is

Option: 1

\frac{2}{3} y


Option: 2

\frac{3}{2} y


Option: 3

\frac{4}{3} y


Option: 4

\frac{3}{4} y


Answers (1)

best_answer

From enthalpy we can see that 2x energy is released by first reaction and only x energy is taken by second reaction and temperature is constant,so second reaction will be 2 times faster than first reaction.

rate_{1}=-\frac{\mathrm{d} \left [ A \right ]}{\mathrm{d} t}= -\frac{1}{3}\frac{\mathrm{d} \left [ B \right ]}{\mathrm{d} t}=\frac{\mathrm{d} \left [ P \right ]}{\mathrm{d} t}

rate_{2}=-\frac{\mathrm{d} \left [ M \right ]}{\mathrm{d} t}= \frac{1}{2}\frac{\mathrm{d} \left [ Q \right ]}{\mathrm{d} t}=\frac{\mathrm{d} \left [ R \right ]}{\mathrm{d} t}

rate_{2}=2\times rate_{1}=\frac{1}{2}\frac{\mathrm{d} \left [ Q \right ]}{\mathrm{d} t}\: \therefore \left [ rate_{1}=\frac{1}{3}y \right ]

2\times \frac{1}{3}y=\frac{1}{2}\frac{\mathrm{d} \left [ Q \right ]}{\mathrm{d} t}

\frac{\mathrm{d} \left [ Q \right ]}{\mathrm{d} t}=\frac{4}{3}y

Hence, the option number (3) is correct.

Posted by

Anam Khan

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