Differential Rate Equation

0.3 M min-1

0.25 M min-1

0.6 M min-1

1.2 M min-1

0.5

0.25

1.0

1.25

days

months

decades

none of these above

0.03 Ms^-1

0.02 Ms^-1

0.017 Ms^-1

0.024 Ms^-1

M s^-1

mol dm^-3 min^-1

M^-1 s^-1

mol L^-1 min^-1

s^-1

Ms^-1

M

M^-1s^-1

0.30 Ms^-1

0.45 Ms^-1

0.20 Ms^-1

0.60 Ms^-1

Both have same rate of depletion.

Rate of depletion of X is doubled from Y.

Rate of depletion of X is halved from Y.

The rate of depletion cannot be determined.

$rate\ =\ -\frac{\text{d}\left[NH_3\right]}{\text{d}t}=-\frac{\text{d}\left[O_2\right]}{\text{d}t}=\frac{\text{d}\left[NO\right]}{\text{d}t}=\frac{\text{d}\left[H_2O\right]}{\text{d}t}$

$rate\ =\ -\frac{1}{2}\frac{\text{d}\left[NH_3\right]}{\text{d}t}=-\frac{1}{3}\frac{\text{d}\left[O_2\right]}{\text{d}t}=\frac{1}{4}\frac{\text{d}\left[NO\right]}{\text{d}t}=\frac{1}{2}\frac{\text{d}\left[H_2O\right]}{\text{d}t}$

$rate\ =\ \frac{\text{d}\left[NH_3\right]}{\text{d}t}=\frac{\text{d}\left[O_2\right]}{\text{d}t}=-\frac{\text{d}\left[NO\right]}{\text{d}t}=-\frac{\text{d}\left[H_2O\right]}{\text{d}t}$

$rate\ =\ -\frac{1}{4}\frac{\text{d}\left[NH_3\right]}{\text{d}t}=-\frac{1}{5}\frac{\text{d}\left[O_2\right]}{\text{d}t}=\frac{1}{4}\frac{\text{d}\left[NO\right]}{\text{d}t}=\frac{1}{6}\frac{\text{d}\left[H_2O\right]}{\text{d}t}$

C2H6 +  $\frac{2}{7}$  O2  $\rightarrow$   $\frac{1}{2}$   CO2 +  $\frac{1}{3}$   H2O

  $\frac{1}{2}$   CO2 +  $\frac{1}{3}$   H2O  $\rightarrow$  C2H6 +  $\frac{2}{7}$  O2

C2H6 +  $\frac{7}{2}$  O2  $\rightarrow$  2 CO2 + 3 H2O

C2H6 +3 O2  $\rightarrow$  2 CO2 + 3 H2O