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Standard XII

Chemistry

Half life

For a reversi...

Question

For a reversible reaction : $N_{2} + O_{2} ⇌ 2 N O$
Activation energy of the backward reaction is lower than that of forward reaction. The slope of k verse 1/T graph will be:

zero

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- \frac{H}{2.303 R}

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\frac{H}{2.303 R}

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- \frac{Δ H}{R}

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Solution

The correct option is C $\frac{H}{2.303 R}$
For reactions whose forward direction has less activation energy, the graph has -ve slope, indicating, a faster reaction in the forward direction.
It is given in the question that the backward reaction has a lower $E_{A}$ than the forward reaction. this means graph has a positive slope.
Slope of graph $= \frac{Δ H}{2.303 R}$
Hence answer is [C].

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For a reversible reaction : N2+O2⇌2NOActivation energy of the backward reaction is lower than that of forward reaction. The slope of k verse 1/T graph will be:

For a reversible reaction : $N_{2} + O_{2} ⇌ 2 N O$
Activation energy of the backward reaction is lower than that of forward reaction. The slope of k verse 1/T graph will be: