For the reaction A(g)⇌B(g)at495KΔrG°=−9.478kJ.If we start the reaction in a closed container at 495K with 22 mill moles of A, the amount of B in the equilibrium mixture is… mill moles (Round off to the Nearest Integer) R=8.31Jmol−1K−1; in 10=2.303
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answer is 20.
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Detailed Solution
Given [A]=22ΔG∘=−9.478−9.478×103=−2.303×8.314×495log[B][A]log[B][A]=9.478×1032.303×8.314×495log[B][A]=94789478log[B][A]=1[B][A]=10x=[A]×10 (x=[B])⇒(22−x)×10=20⇒(22−x)=2x=20