the following equilibrium reaction is investigated. 2N2O(g) + N2H4(g) --- 3N2(g)

Question

the following equilibrium reaction is investigated. 2N2O(g) + N2H4(g) --- 3N2(g) + 2H2O(g) when 0.10 moles of N2O and 0.25 moles of N2H4 are placed in a 10.0L container and allowed to come to equilibrium, the equilibrium concentration of N2O is 0.0060M. What is the equilibrium concentration of all four substances? the following equilibrium reaction is investigated. 2N2O(g) + N2H4(g) --- 3N2(g) + 2H2O(g) when 0.10 moles of N2O and 0.25 moles of N2H4 are placed in a 10.0L container and allowed to come to equilibrium, the equilibrium concentration of N2O is 0.0060M. What is the equilibrium concentration of all four substances? 2N2O(g) + N2H4(g) --- 3N2(g) + 2H2O(g) when 0.10 moles of N2O and 0.25 moles of N2H4 are placed in a 10.0L container and allowed to come to equilibrium, the equilibrium concentration of N2O is 0.0060M. What is the equilibrium concentration of all four substances?

Explanation / Answer

2N2O(g) + N2H4(g) --- 3N2(g) + 2H2O(g)

when 0.10 moles of N2O and 0.25 moles of N2H4 are placed in a 10.0L container and allowed to come to equilibrium, the equilibrium concentration of N2O is 0.0060M.

It means 0.0060*10 moles or 0.06 moles of N2O is left over.

or 0.10 - 0.06 moles or 0.04 mole of N2O was converted

                               2N2O(g) + N2H4(g) --- 3N2(g) + 2H2O(g)

Initial concentration   0.10           0.25          0            0

amount used/produced 0.04        0.02         0.06      0.04

final concentration       0.06           0.23        0.06      0.04

equilibrium concentration of NO2 will be 0.06/10 or 0.006 M

equilibrium concentration of N2H4 will be 0.23/10 or 0.023 M

equilibrium concentration of N2 will be 0.06/10 or 0.006 M

equilibrium concentration of H2O will be 0.04/10 or 0.004 M

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