Question

step by step please. Write neat please. Read the direction carefully please. Box in answer please. use formula sheet

3. Consider the reaction: Suppose that, at a particular moment during the reaction, hydrogen gas is beling H2 ()he equation fat (1 o PHx) P() formed at the rate of 0.078 M/s A) Express the rate of the reaction in terms of the change in concentration or each reactant and product in the box below (1 pt) B) At what rate is Paig) being formed? (1 pt) Answer At what rate is PHs ( reacting? (1 pt) c) Answer In[A]- In[A]o-kt [A]e In [A]o GEN CHEM II HANDOUT h 6.626 x 1034 J-sec -kt c 3.00 x 108 m/sec Kp Ke(RT)an PU=nRT kg-m2 J pH--log (H) R-0O21atm mol-K L s2 p0H 14.00 [A] [Alo-kt 0.693 pH R-8.314 t1 mol-K k -bt vb-4ac hc AE h-v- C= As K-Ks= Kw 2a E=mc kr [Alo ti= AE = Am J C-V 2k Ea In- 1 Ink=a InA RT RT1T 1 kz In ks 1 mole e-96,500C K "C+273 R Ea k= A.e RT 1 AS EndSoduct2mAS (reactants) + kt [A]o [A] Am-(mass of nucleus)-(mass of nuclear components) AG=AHsys-TASsys Ecall = E 0.0592V -log Q AG=AG+RT In Q pH pK+loebase lacid n AG=-RT In K E cell inQ Eo-0.0257V 0.0257V E'cell In K 0.0592V E log K AG= -nFEcell cell Kw= [H*][OH] 71 1.0 x 1014 CC/CHEM1118/F2018/HANDOUT Ecell E"cathode- E anode

Answer 1

Given reaction :-
+P4(a)H2(g)

The balanced reaction is  
PA(g)6H2(g) APH3(9)

(A) The rate of reaction is

ΙΔΡΗ] Averagerateof ConsumptionofPH3 At

Averagerateof formationofP = +F| At

Averagerateof formationof H= +1A[H2] 6 At

$Average rate of reaction =-\frac{1}{4} \frac{\Delta[PH_{3}]}{\Delta t} = + \frac{\Delta [P_{4}]}{\Delta t} = + \frac{1}{6} \frac{\Delta[H_{2}]}{\Delta t}$

Given:-
AH2 0.078M/s At
.

(B) rate of formation of P₄ :

we have,

$+ \frac{\Delta [P_{4}]}{\Delta t} = + \frac{1}{6} \frac{\Delta[H_{2}]}{\Delta t}$

APA+( x 0.078M/s + At 6

$+ \frac{\Delta [P_{4}]}{\Delta t} = + 0.013 M/s$

(C) rate of consumption of PH₃

we have,

$-\frac{1}{4} \frac{\Delta[PH_{3}]}{\Delta t} = + \frac{1}{6} \frac{\Delta[H_{2}]}{\Delta t}$

$\frac{\Delta[PH_{3}]}{\Delta t} = - \frac{4}{6} \frac{\Delta[H_{2}]}{\Delta t}$

$\frac{\Delta[PH_{3}]}{\Delta t} = - (\frac{4}{6}) \times 0.013 M/s$

$\frac{\Delta[PH_{3}]}{\Delta t} = - 0.052 M/s$

Note:-

Rate of reaction:- Rate of reaction is definded as the change in concentration of reactant or productper unit time.

Average rate of reaction:-

it is defined as the change in concentration of reactant or product divided by the time interval over which the change is occurs.

$Average rate = \frac{change in concentration of species}{chang in time} = \frac{\Delta C}{\Delta t}$

Consider a reaction:- $A \rightarrow B$

$Average rate of consumption of A = - \frac{\Delta [A]}{\Delta t}$

$Average rate of formation of B = + \frac{\Delta [B]}{\Delta t}$

Hence, $Average rate of reaction =-\frac{\Delta[A]}{\Delta t} = + \frac{\Delta [B]}{\Delta t}$

​​​​​​For reactions involving different number of moles of substances the various reactants are consumed at different rates and various products are formed at different rates.

For example.

N2(g)3H2g)2NH3)
....(A)

here there are three different rates. However the rate of reaction must unique and must be same no matter which reactant or product is chosen to measure. To avoid the existence of multiple rates, the general rate of reaction is equal to the rate of consumption of a reactant or the rate of formation of the product divided by its coefficient in the balanced chemical equation.

Thus for reaction A.

$Average rate of reaction =-\frac{\Delta[N_{2}]}{\Delta t} = -\frac{1}{3} \frac{\Delta [H_{2}]}{\Delta t} = + \frac{1}{2} \frac{\Delta[NH_{3}]}{\Delta t}$