Variation of log 10 K with 1 T is shown by the graph in which straight line is at 45 o C- hence - H o is-

The correct option is B 19.147kJ/mol Slope of reaction is nbsp; m = tanθ = 1 = Δ H^0/2.303R; nbsp;Δ H^0 = 2.303*R = 19.11kJ/mol ...

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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

Variation of ...

Question

Variation of ${log}_{10} K$ with $\frac{1}{T}$ is shown by the graph in which straight line is at $45^{o} C$ , hence ${Δ H}^{o}$ is:

- 4.606 k J / m o l

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- 19.147 k J / m o l

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8.314 k J / m o l

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10 k J / m o l

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Solution

The correct option is B $- 19.147 k J / m o l$
Slope of reaction is
$m = t a n θ = 1 = - Δ H^{0} / 2.303 R; Δ H^{0} = - 2.303 * R = - 19.11 k J / m o l$

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Similar questions

Q. Variation of

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Variation of log10K with 1T is shown by the graph in which straight line is at 45oC, hence ΔHo is:

The correct option is B −19.147kJ/molSlope of reaction is m=tanθ=1=−ΔH0/2.303R;ΔH0=−2.303∗R=−19.11kJ/mol

Variation of ${log}_{10} K$ with $\frac{1}{T}$ is shown by the graph in which straight line is at $45^{o} C$ , hence ${Δ H}^{o}$ is:

The correct option is B $- 19.147 k J / m o l$
Slope of reaction is
$m = t a n θ = 1 = - Δ H^{0} / 2.303 R; Δ H^{0} = - 2.303 * R = - 19.11 k J / m o l$