Consider the follwing first order reaction:--------- 4PH3---> P4 + 6H2----------

Question

Consider the follwing first order reaction:--------- 4PH3---> P4 + 6H2---------- The half-life of the reaction is 35 seconds at 680 degrees Celcius. Calculate the rate constant for the reaction , and the time required for 95% of the PH3 to decompose.--------- 8.) The rate constant for the second order reaction below is .54 M^-1 * s^-1. How long in seconds would it take for the concentration of NO2 to decrease from .65 M to .18 M? ------------2NO2---->2NO+O2 Show all steps with correct answer for 5 star rating. Thank you.

Explanation / Answer

FOLLOW THIS We actually do not have to look at the rates. We must simply understand that if hydrogen is formed, then PH3 is being used up or reacted and a certain amount (moles) of it is used up as well. And if at a certain point in time when a particular amount of hydrogen is produced over a period of time, then the fixed or corresponding amount of reactants (PH3) is used as well. We must understand that for 6 H2 to form, we must use 4 PH3. This would be how i would attempt it (its my first time seeing rates with moles ) no.of moles of H2 > 0.078 moles. molar ratio of PH3 to H2 > 4 : 6 = 2 : 3 no. of moles of PH3 used > 0.078 divided by 3 then multiplied by 2. = 0.052 From that we know that when 0.078 moles of Hydrogen is produced per second, 0.052 moles of PH3 MUST BE used per second. Similarly for rate of production of P4 : molar ratio of P4 to H2 > 1 : 6 no. of moles of P4 > 6 x 0.078 = x (i do not have a calculator >

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