Answer to Question #229110 in General Chemistry for Mickayla Ennis

Question #229110

Calculate the reaction free energy of

H2(g) + Cl2(g)   2HCl(g)

when the concentrations are 0.026 mol L-1 (H2), 0.033 mol L-1 (Cl2), and 0.00184 mol L-1 (HCl), and the temperature is 500K. For this reaction Kc = 85 at 500 K.

  1. -5.3 kJ
  2. -12.2 kJ
  3. -22.2 kJ
  4. -41.5 kJ




1
Expert's answer
2021-08-30T02:26:24-0400

Reaction quotient

"Q_c = \\frac{[HCl]^2}{[H_2][Cl_2]} \\\\\n\n= \\frac{0.00184^2}{(0.026 \\times 0.033)}=0.003946 \\\\\n\nK_c=85 \\\\\n\nT= 500 \\;K \\\\\n\nR=8.314 \\;J\/mol \\cdot K \\\\\n\n\u0394G=\u0394G_o +Rtln(Q_c) \\\\\n\n= -RT ln(K_c) + RT ln(Q_c) \\\\\n\n= -8.314 \\times 500 \\times ln(85) + 8.314 \\times 500 \\times ln(0.003946) \\\\\n\n= -41477 \\;J\/mol \\\\\n\n\u2248 -41.5 \\;kJ\/mol"

Answer: the last option -41.5 kJ/mol


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