Question

The following reaction is a slow reaction and hence Ruthenium (molar mass = 101gm) is used as catalyst

$NaBH_4(aq) + 2H_2O(l) \xrightarrow{Catalyst} Na^+(aq) + BO_2^- + (aq) + 4H_2(g)$

Kinetic studies show that reaction is 1st order with respect to catalyst but zero-order with respect to substrate. The rate of hydrogen production per mole of Ruthenium is 92 mol $H_2(molRu)^{-1}min^{-1}$ at $27°C$. Identify CORRECT option(s) {R = 0.08 atm - $l/mol-K$}

• A. Approximately 0.045 mgRu, must be added to 0.1 $dm^3$, 1 $moldm^{-3}NaBH_4$ to supply $H_2$ gas at rate of 0.1 $dm^3min^{-1}$ at $27°C$ and 1 atm
• B. Approximately 4.574 mg of Ru, must be added to 0.1 $dm^3$, 1 $moldm^{-3}NaBH_4$ to supply $H_2$ gas at rate of 0.1 $dm^3min^{-1}$ at $27°C$ and 1 atm
• C. If 0.1 $dm^3$, 1 $moldm^{-3}NaBH_4$ is used to supply $H_2$ at at rate of 0.1 $dm^3min^{-1}$ at $27°C$, 1 atm, then upto 96 min hydrogen can be supplied in this way
• D. If 0.1 $dm^3$, 1 $moldm^{-3}NaBH_4$ is used to supply $H_2$ at at rate of 0.1 $dm^3min^{-1}$ at $27°C$, 1 atm, then upto 48 min hydrogen can be supplied in this way

Answer 1

Answer: (B), (C)

Approximately 4.574 mg of Ru, must be added to 0.1 $dm^3$, 1 $moldm^{-3}NaBH_4$ to supply $H_2$ gas at rate of 0.1 $dm^3min^{-1}$ at $27°C$ and 1 atm

If 0.1 $dm^3$, 1 $moldm^{-3}NaBH_4$ is used to supply $H_2$ at at rate of 0.1 $dm^3min^{-1}$ at $27°C$, 1 atm, then upto 96 min hydrogen can be supplied in this way

Answer 2

The reaction is given by: $NaBH_4(aq) + 2H_2O(l) \xrightarrow{Catalyst} Na^+(aq) + BO_2^- (aq) + 4H_2(g)$

The rate law is first order with respect to the catalyst (Ruthenium) and zero order with respect to the substrate ($NaBH_4$). Rate of reaction = $k [Ru]^1 [NaBH_4]^0 = k [Ru]$.

The problem states that the rate of hydrogen production per mole of Ruthenium is 92 mol $H_2(molRu)^{-1}min^{-1}$. Let $n_{Ru}$ be the number of moles of Ruthenium catalyst. The rate of hydrogen production, $R_{H_2}$, is given by: $R_{H_2} = 92 \text{ mol } H_2 (molRu)^{-1}min^{-1} \times n_{Ru} \text{ mol Ru} = 92 n_{Ru} \text{ mol } H_2 min^{-1}$.

We are given the desired rate of $H_2$ gas supply as 0.1 $dm^3 min^{-1}$ at $27°C$ and 1 atm. Using the ideal gas law, $PV=nRT$, we can convert this volume rate to a molar rate. $V = 0.1 \, dm^3 = 0.1 \, L$ $P = 1 \, atm$ $T = 27°C = 27 + 273 = 300 \, K$ $R = 0.08 \, atm \cdot L / (mol \cdot K)$

The number of moles of $H_2$ per minute is: $n_{H_2} = \frac{PV}{RT} = \frac{(1 \, atm) \times (0.1 \, L)}{(0.08 \, atm \cdot L / (mol \cdot K)) \times (300 \, K)} = \frac{0.1}{24} \, mol = \frac{1}{240} \, mol$. So, the desired rate of $H_2$ production is $R_{H_2} = \frac{1}{240} \, mol \, H_2 min^{-1}$.

Now we equate this to the expression for $R_{H_2}$ in terms of $n_{Ru}$: $92 n_{Ru} = \frac{1}{240}$ $n_{Ru} = \frac{1}{240 \times 92} \, mol$.

To find the mass of Ruthenium, we use its molar mass (101 g/mol): Mass of Ru = $n_{Ru} \times \text{Molar mass of Ru} = \frac{1}{240 \times 92} \, mol \times 101 \, g/mol = \frac{101}{22080} \, g$.

Let's calculate this value in milligrams: Mass of Ru $= \frac{101}{22080} \times 1000 \, mg = \frac{101000}{22080} \, mg = \frac{10100}{2208} \, mg \approx 4.574 \, mg$. Approximately 4.574 mg of Ru is needed.

Now let's consider the duration for which hydrogen can be supplied. We are given 0.1 $dm^3$ (0.1 L) of 1 $moldm^{-3}$ (1 mol/L) $NaBH_4$ solution. The initial amount of $NaBH_4$ is: Moles of $NaBH_4 = \text{Volume} \times \text{Concentration} = (0.1 \, L) \times (1 \, mol/L) = 0.1 \, mol$.

From the reaction stoichiometry, 1 mole of $NaBH_4$ produces 4 moles of $H_2$. The total amount of $H_2$ that can be produced from 0.1 mol of $NaBH_4$ is: Total moles of $H_2 = 0.1 \, mol \, NaBH_4 \times \frac{4 \, mol \, H_2}{1 \, mol \, NaBH_4} = 0.4 \, mol \, H_2$.

The desired rate of $H_2$ production is constant at $\frac{1}{240} \, mol/min$, as long as the required amount of catalyst is present and $NaBH_4$ is available. Since the reaction is zero-order with respect to $NaBH_4$, the rate per mole of catalyst is constant as long as $NaBH_4$ is present. With a fixed amount of catalyst calculated above, the total rate of $H_2$ production is constant. Let $t_{max}$ be the maximum time in minutes for which $H_2$ can be supplied at this rate. The total moles of $H_2$ produced in time $t_{max}$ is (rate of $H_2$ production) $\times t_{max}$. This total amount cannot exceed the total amount of $H_2$ that can be produced from the initial $NaBH_4$. $(\frac{1}{240} \, mol/min) \times t_{max} = 0.4 \, mol$ $t_{max} = 0.4 \times 240 \, minutes = \frac{4}{10} \times 240 \, minutes = \frac{2}{5} \times 240 \, minutes = 2 \times 48 \, minutes = 96 \, minutes$.

So, hydrogen can be supplied at the desired rate for upto 96 minutes.