A student ran the following reaction in the laboratory at 291 K: 2CH2Cl2(g) CH4(g) + CCl4(g)...

A student ran the following reaction in the laboratory at 291 K: 2CH2Cl2(g) CH4(g) + CCl4(g) When she introduced 6.63×10-2 moles of CH2Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of CH2Cl2(g) to be 4.92×10-3 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.

Kc = _____

Expert Solution

queen_honey_blossom answered 2 years ago

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A student ran the following reaction in the laboratory at 445 K : PCl5(g)-------> PCl3(g) + Cl2(g) When she introduced 1.28 moles of PCl5(g) into a 1.00 liter container, she found the equilibrium concentration of PCl5(g) to be 1.24 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.

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The equilibrium constant, K, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) =CH4(g) +...

The equilibrium constant, K, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) =CH4(g) + CCl4(g) An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.12×10-2 M CH2Cl2, 0.166 M CH4 and 0.166 M CCl4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.121 mol of CH4(g) is added to the flask? [CH2Cl2] = _________M [CH4] = _________ M [CCl4] = _________ M

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A student ran the following reaction in the laboratory at 610 K: CO(g) + Cl2(g) COCl2(g)...

A student ran the following reaction in the laboratory at 610 K: CO(g) + Cl2(g) COCl2(g) When she introduced 0.183 moles of CO(g) and 0.211 moles of Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of Cl2(g) to be 6.72×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. 2.A student ran the following reaction in the laboratory at 546 K: COCl2(g) CO(g) + Cl2(g) When she introduced 0.854 moles of COCl2(g) into a 1.00...

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