Factories the following polynomial : a) 5m² - 6m
Answers
Step-by-step explanation:
Zero order reaction:
\rm A(g) \longrightarrow B(g) + C(g)A(g)⟶B(g)+C(g)
Completion time for the zero order reaction (t) = 13.86 min
Half life of zero order reaction:
\begin{gathered} \rm {(t_{1/2})}_{zero \: order} = \dfrac{t}{2} \\ \\ \rm {(t_{1/2})}_{zero \: order} = \dfrac{13.86}{2} \\ \\ \rm {(t_{1/2})}_{zero \: order} = 6.93\end{gathered}
(t
1/2
)
zeroorder
=
2
t
(t
1/2
)
zeroorder
=
2
13.86
(t
1/2
)
zeroorder
=6.93
First order reaction:
\rm X(g) \longrightarrow Y(g)X(g)⟶Y(g)
Half life of first order reaction:
\rm {(t_{1/2})}_{first \: order} = \dfrac{ln 2}{k}(t
1/2
)
firstorder
=
k
ln2
According to the question, Half life for the zero order reaction and half life for the first order reaction are equal.
\begin{gathered} \rm \leadsto {(t_{1/2})}_{zero \: order} = {(t_{1/2})}_{first \: order} \\ \\ \rm \leadsto 6.93 = \dfrac{ln2}{k} \\ \\ \rm \leadsto k = \dfrac{ln2}{6.93} \\ \\ \rm \leadsto k = \dfrac{0.693}{6.93} \\ \\ \rm \leadsto k = 0.1 \: min^{ - 1} \\ \\ \rm \leadsto k = 6 \: hr^{ - 1}\end{gathered}
⇝(t
1/2
)
zeroorder
=(t
1/2
)
firstorder
⇝6.93=
k
ln2
⇝k=
6.93
ln2
⇝k=
6.93
0.693
⇝k=0.1min
−1
⇝k=6hr
−1
\therefore∴ Rate constant for reaction X(g) → Y(g) = 6 \sf hr^{-1}hr
−1
Answer:
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Step-by-step explanation:
5m2 - 6m
5m2 - 6m = m • (5m - 6)
m • (5m - 6)