For the reaction 4NH3 + 5O2 → 4NO + 6H2O, if the rate of disappearance of NH3 is 3.6 × 10-3 mol L-1 s-1, what is the rate of formation of H2O ?

Question

For the reaction 4NH3 + 5O2 → 4NO + 6H2O, if the rate of disappearance of NH3 is 3.6 × 10-3 mol L-1 s-1, what is the rate of formation of H2O ? (A) 5.4 × 10-3 mol L-1 s-1 (B) 3.6 × 10-3 mol L-1 s-1 (C) 4× 10-4 mol L-1 s-1 (D) 0.6 × 10-4 mol L-1 s-1

Tardigrade Admin · Accepted Answer

-(1/4)(d[NH3]/dt)=+(1/6) (d [H2O]/dt) (d[H2O]/dt)=(6/4)×3.6×10-3 =5.4×10-3 mol L-1 s-1

Q. For the reaction $4 N H_{3} + 5 O_{2} \to 4 NO + 6 H_{2} O$ , if the rate of disappearance of $N H_{3}$ is $3.6 \times 1 0^{- 3} m o l L^{- 1} s^{- 1}$ , what is the rate of formation of $H_{2} O$ ?

$5.4 \times 1 0^{- 3} m o l L^{- 1} s^{- 1}$

$3.6 \times 1 0^{- 3} m o l L^{- 1} s^{- 1}$

$4 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}$

$0.6 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}$

$- \frac{1}{4} \frac{d [ N H _{3} ]}{d t} = + \frac{1}{6} \frac{d [ H _{2} O ]}{d t}$
$\frac{d [ H _{2} O ]}{d t} = \frac{6}{4} \times 3.6 \times 1 0^{- 3}$
$= 5.4 \times 1 0^{- 3} m o l L^{- 1} s^{- 1}$

Q. For the reaction 4NH3​+5O2​→4NO+6H2​O, if the rate of disappearance of NH3​ is 3.6×10−3molL−1s−1, what is the rate of formation of H2​O ?

5.4×10−3molL−1s−1

3.6×10−3molL−1s−1

4×10−4molL−1s−1

0.6×10−4molL−1s−1