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Heat of Reaction

A graph plott...

Question

A graph plotted between ${log}_{10} K_{c}$ and $1 / T$ is a straight line with intercept $10$ and slope equal to $0.5$ . Calculate heat of reaction at $298 K$ .

Δ H = - 2.303 c a l m o l^{- 1}

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Δ H = 2.303 c a l m o l^{- 1}

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Δ H = - 4.606 c a l m o l^{- 1}

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Δ

H = - 6.909

c a l m o l^{- 1}

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Solution

The correct option is B $Δ H = - 2.303 c a l m o l^{- 1}$
Solution:- (A) $- 2.303 c a l / m o l$
As we know that,
The slope of the straight line plotted between ${log}_{10} K_{c}$ and $\frac{1}{T}$ will be given as-
$m = - \frac{Δ H}{2.303 R}$
The slope of the line is given as $0.5$ .
$∴ - \frac{Δ H}{2.303 R} = 0.5$
$\Rightarrow Δ H = - 2.303 \times 2 \times 0.5$
$\Rightarrow Δ H = - 2.303 c a l / m o l$
Hence the heat of reaction at $298 K$ is $- 2.303 c a l / m o l$ .

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Similar questions

Q. Assertion :As temperature increases, heat of reaction also increases for exothermic as well as for endothermic reactions. Because Reason:

Δ

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{Δ H}_{2} (a t T_{2}) = {Δ H}_{1} (a t T_{1}) + {Δ C}_{P} (T_{2} - T_{1})

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The mean heat capacities over this temperature range are,

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A graph plotted between log10Kc and 1/T is a straight line with intercept 10 and slope equal to 0.5. Calculate heat of reaction at 298 K.

A graph plotted between ${log}_{10} K_{c}$ and $1 / T$ is a straight line with intercept $10$ and slope equal to $0.5$ . Calculate heat of reaction at $298 K$ .