HSC Science (Computer Science) 12th Board ExamMaharashtra State Board
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The equilibrium constant Kp for the reaction, H2(g) + I2(g) → 2HI(g) is 130 at 510 K. - HSC Science (Computer Science) 12th Board Exam - Chemistry

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Question

The equilibrium constant Kp for the reaction,

H2(g) + I2(g) → 2HI(g) is 130 at 510 K. Calculate ΔGo for the following reaction at the same temperature: 2HI(g) → H2(g) + I2(g) [Given: R = 8.314 J K-1 mol-1 ]

Solution

Given: H2(g) + I2(g) ⇌ 2HI(g); Kp = 130, T = 510 K, R = 8.314 J K-1 mol-1

To find: ΔGº for reaction, 2HI(g) ⇌ H2(g) + I2(g)

Formula: ΔGº = -2.303 RT log10Kp

Calculation:

H2(g) + I2(g) ⇌ 2HI(g); Kp = 130,

2HI(g)⇌H2(g) + I2(g) ; Kp = 1/130

ΔGº= - 2.303 RT log10Kp

=-2.303 x 8.314 x 510 x log10 (1/130)

= 20642.7 J mol-1
=20.64 kJ mol-1

  Is there an error in this question or solution?

APPEARS IN

 2014-2015 (October) (with solutions)
Question 4.2 | 7.00 marks
Solution The equilibrium constant Kp for the reaction, H2(g) + I2(g) → 2HI(g) is 130 at 510 K. Concept: Chemical Thermodynamics and Energetic - Equilibrium Constant.
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