For the reaction H 2 g - I 2 g - 2 HI g K c -66-9 at 350- C and K c -50-0 at 448- C the reaction hasA- - H - veB- - H - veC- H - zeroD- H sign can not be determined

The correct option is B H = veH_2(g) + I_2 (g) ⇋ 2HI(g) logK_2K_1 = H2.303R[1T_1 1T_2 ] Given K_1 = 66.9 T_1 = 350^∘C or 623 K K_2 = 50 T_2 = 448^∘C or ...

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Byju's Answer

Standard XII

Chemistry

Valency and Oxidation State

For the react...

Question

For the reaction
$H_{2} (g) + I_{2} (g) ⇋ 2 H I (g)$
$K_{c} = 66.9$ at $350^{\circ} C$ and $K_{c} = 50.0$ at $448^{\circ} C$ the reaction has

△ H = + v e

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△ H = - v e

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△ H = z e r o

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△ H

sign can not be determined

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Solution

The correct option is B $△ H = - v e$
$H_{2} (g) + I_{2} (g) ⇋ 2 H I (g) log \frac{K_{2}}{K_{1}} = \frac{△ H}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
Given $K_{1} = 66.9$
$T_{1} = 350^{\circ} C o r 623 K K_{2} = 50 T_{2} = 448^{\circ} C o r 721 K log \frac{50}{66.9} = \frac{△ H}{2.303 R} [\frac{1}{623} - \frac{1}{721}]$
After calculation negative value of $△ H$ is obtained.

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For the reaction H2(g)+I2(g)⇋2HI(g) Kc=66.9 at 350∘C and Kc=50.0 at 448∘C the reaction has

The correct option is B △H=−veH2(g)+I2(g)⇋2HI(g)logK2K1=△H2.303R[1T1−1T2] Given K1=66.9 T1=350∘Cor623KK2=50T2=448∘Cor721Klog5066.9=△H2.303R[1623−1721] After calculation negative value of △H is obtained.

For the reaction
$H_{2} (g) + I_{2} (g) ⇋ 2 H I (g)$
$K_{c} = 66.9$ at $350^{\circ} C$ and $K_{c} = 50.0$ at $448^{\circ} C$ the reaction has