Part A Consider the combustion of ethylene: C3H1(g) + 302 (g)2CO2 (g) + 2H2 0(g)

Question

Part A Consider the combustion of ethylene: C3H1(g) + 302 (g)2CO2 (g) + 2H2 0(g) If the concentration of C2 H4 is decreasing at the rate of 3.8x102 M/s, what is the rate of change in the concentration of CO2 Express the rate in molarity per second to two significant digits. ? M/s Submit Previous Answers Request Answer X Incorrect; Try Again In the balanced chemical equation, every mole of C2 Hs reacts to form two moles of CO2. Use that stoichiometric relationship to determine the rate of change for CO2

Explanation / Answer

Part-A

C2H4(g) + 3 O2(g) --------> 2 CO2(g) + 2 H2O(g)

1 mole C2H4 on combustion produce 2 moles CO2 and 2 moles H2O

Rate of decreasing concentration of C2H4 =3.8 x 10-2 M/s

Rate of increasing concentration of CO2 = 2 x 3.8 x 10-2 = 7.6 x 10-2 M/s = 0.076 M/s

Part-B

Rate of increasing concentration of H2O = 2 x 3.8 x 10-2 = 7.6 x 10-2 M/s = 0.076 M/s

Part-C

                                    N2H4(g) + H2(g) ------> 2 NH3(g)

1 mole of N2H4 reacts with 1 mole of H2 to give 2 moles of NH3

Rate of decrease in the partial pressure of N2H4 = 74 torr/hour

Rate of increase in the partial pressure of NH3 = 2 x 74 = 148 torr/hour = 1.48 x 102 torr/hour

Part-D

                                     N2H4(g) + H2(g) ------> 2 NH3(g)

In the above reaction, the total number of moles of gaseous reactants (1+1=2) is equal to number moles of gaseous products (2). Hence, there is no change in the total pressure in the vessel during the course of reaction.

Rate of change of total pressure in the vessel during the course of reaction = 0.0 torr/hour

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