For a hypothetical reaction, A + 3B → P \DeltaH = –2 x kJ/mol M → 2Q + R \DeltaH = + x kJ/mol 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

Engineering

Chemistry

Rate of Chemical Reaction

Question

For a hypothetical reaction,
A + 3B → P $∆$ H = –2 x kJ/mol

M → 2Q + R $∆$ H = + x kJ/mol
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 :

$\frac{2}{3} y$

$\frac{3}{2} y$

$\frac{4}{3} y$

$\frac{3}{4} y$

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To find the rate of formation of Q, we relate it to the rate of disappearance of B using stoichiometry. For the first reaction A + 3B → P, the rate of disappearance of B is given as y M/s. The stoichiometric coefficient for B is 3, so the rate of this reaction, r₁, is (1/3) * (rate of B disappearance) = y/3.

The second reaction M → 2Q + R is endothermic, but since the temperature is constant, an equal magnitude of heat is absorbed from the first exothermic reaction, implying the rates of the two reactions are coupled to maintain energy balance. The enthalpy change for the first reaction is -2x kJ/mol and for the second is +x kJ/mol. For the heat effects to cancel (ΔH_total=0), the rate of the second reaction, r₂, must be twice that of the first: r₂ = 2 * r₁ = 2*(y/3) = 2y/3.

The rate of formation of Q is related to r₂ by its stoichiometric coefficient (2): rate of Q formation = 2 * r₂ = 2 * (2y/3) = 4y/3 M/s.

Final Answer: $\frac{4}{3} y$