For chemical reactions involving ideal gases, the equilibrium constant K can be
ID: 683564 • Letter: F
Question
For chemical reactions involving ideal gases, the equilibrium constant K can be expressed either in terms of the concentrations of the gases (in M) or as a function of the partial pressures of the gases (in atmospheres). In the latter case, the equilibrium constant is denoted as Kp to distinguish it from the concentration-based equilibrium constant K. For the reaction 2CH4(g) C2H2(g) + 3H2(g) K = 0.140 at 1567 degree C. What is Kp for the reaction at this temperature? Express your answer numerically. For the reaction N2(g) + 3H2(g) 2NH3(g) Kp = 2.45 times 10-3 at 337 degree C. What is K for the reaction at this temperature? Enter your answer numerically.Explanation / Answer
We Know that : The given Reaction is : 2 CH4 (g ) <-------> C2H2 ( g) + 3 H2 ( g ) Kp = Kc ( RT )ng ng = gaseous moles of products - gaseous moles of reactants = 4 - 2 = 2 Kp = 0.140 ( 0.0821 x 1840 K )2 = 3194.8464 b. N2 ( g ) + 3 H2 ( g) <---------> 2 NH3 ( g ) Kp = Kc ( RT )ng ng = gaseous moles of products - gaseous moles of reactants = 2 - 4 = - 2 2.45 x 10-3 = Kc ( 0.0821 x 610)-2 Kc = 6.1448 ng = gaseous moles of products - gaseous moles of reactants = 4 - 2 = 2 Kp = 0.140 ( 0.0821 x 1840 K )2 = 3194.8464 b. N2 ( g ) + 3 H2 ( g) <---------> 2 NH3 ( g ) Kp = Kc ( RT )ng ng = gaseous moles of products - gaseous moles of reactants = 2 - 4 = - 2 2.45 x 10-3 = Kc ( 0.0821 x 610)-2 Kc = 6.1448 Kc = 6.1448Related Questions
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