Q.How many mL of 0.1 M HCl are required to react completely with 1 g mixture of $Na_2CO_3$ and $NaHCO_3$ containing equimolar amounts of both?

The problem gives a rate of disappearance of hydrogen and asks for the amount of ammonia formed in a different time interval. The central concept is stoichiometric proportionality in chemical kinetics: for a balanced reaction, the rates of consumption and production of species are related by their coefficients. Once we know how fast hydrogen is used up, we can directly find how fast ammonia appears, then multiply by the new time.

1. Write the balanced equation The synthesis of ammonia is:

This tells us: for every 3 moles of $\mathrm{H_2}$ consumed, 2 moles of $\mathrm{NH_3}$ are produced.

2. Find the rate of hydrogen disappearance Given: $6 \times 10^{-2}$ moles of $\mathrm{H_2}$ disappear in 10 minutes. So the rate of disappearance of $\mathrm{H_2}$ is:

3. Use stoichiometry to find the rate of ammonia formation From the balanced equation:

Therefore:

4. Calculate moles of ammonia formed in 0.3 minutes $$$$

Step 1 — Why the half-life shortcut is legitimate here.

For a first-order reaction the integrated rate law is

and the half-life

is independent of the initial concentration. That is the special feature of first order: every successive $20\,\mathrm{s}$ window halves whatever is present at the start of that window.

Step 2 — Count the half-lives.

Step 3 — Halve four times.

Tracking it explicitly:

Concept & Intuition

For a reaction $aA \to bB$, the rate is defined consistently for every species:

This holds even when the coefficients are fractions, as here: $\dfrac{3}{5}X \to \dfrac{1}{2}Y + Z$.

Step-by-step

1. Define the common rate $r$.

   For X (coefficient $3/5$): $r = -\dfrac{1}{3/5}\dfrac{d[X]}{dt} = -\dfrac{5}{3}\dfrac{d[X]}{dt}$.

   For Y (coefficient $1/2$): $r = +\dfrac{1}{1/2}\dfrac{d[Y]}{dt} = +2\dfrac{d[Y]}{dt}$.

2. Equate.

3. Solve for $-d[X]/dt$. …

The key idea is that for a reaction

the rate of reaction $r$ is defined as

The negative sign is used because reactant concentrations decrease over time, making their derivatives negative, so the rate comes out positive. For products, the sign is positive.

Here, the reactants are $X$, $Y$, and $Z$ with coefficients $3$, $2$, and $\frac{5}{2}$ respectively. So the rate must satisfy:

We then rewrite these equalities in the form given in the options.

1. Write the rate equalities from definition

(since $1/(5/2) = 2/5$). The negatives cancel, so we have:

2. Convert to the form in the options Options express relationships like $k_1 \frac{d[X]}{dt} = k_2 \frac{d[Y]}{dt} = k_3 \frac{d[Z]}{dt}$. From $\frac{1}{3}\frac{d[X]}{dt} = \frac{1}{2}\frac{d[Y]}{dt}$, multiply both sides by 6:

From $\frac{1}{3}\frac{d[X]}{dt} = \frac{2}{5}\frac{d[Z]}{dt}$, multiply both sides by 15:

So one valid set is:

3. Match to the given options We need to find which option’s equalities are equivalent to $\frac{1}{3}\frac{d[X]}{dt} = \frac{1}{2}\frac{d[Y]}{dt} = \frac{2}{5}\frac{d[Z]}{dt}$. Check option (C):

Divide each term by its coefficient to see the common rate:

• From first equality: $\frac{4}{15}\frac{d[X]}{dt} = \frac{4}{5}\frac{d[Y]}{dt}$ → multiply by 15/4: $\frac{d[X]}{dt} = 3\frac{d[Y]}{dt}$? Wait, let’s do carefully:

 But from our definition, $\frac{1}{3}\frac{d[X]}{dt} = \frac{1}{2}\frac{d[Y]}{dt}$ gives $\frac{d[X]}{dt} = \frac{3}{2}\frac{d[Y]}{dt}$, not $3$. So this seems off? Let’s re-check systematically.

Actually, the correct method: For option (C), set the common value as $k$. Then:

Now compute $-\frac{1}{3}\frac{d[X]}{dt} = -\frac{1}{3}\cdot\frac{15}{4}k = -\frac{5}{4}k$.

$-\frac{1}{2}\frac{d[Y]}{dt} = -\frac{1}{2}\cdot\frac{5}{4}k = -\frac{5}{8}k$. These are not equal — so (C) fails? That suggests I made a sign error: Actually, the rate definition uses negative of reactant derivatives. Let’s re-derive carefully.

4. Correct derivation with signs

Multiply each equality by the denominators to eliminate fractions:

From $-\frac{1}{3}\frac{d[X]}{dt} = -\frac{1}{2}\frac{d[Y]}{dt}$, multiply by $-6$:

From $-\frac{1}{3}\frac{d[X]}{dt} = -\frac{2}{5}\frac{d[Z]}{dt}$, multiply by $-15$:

Step 1 — Oxidation half-reaction (reduction of permanganate in basic medium).

In $\mathrm{MnO_4^-}$, Mn is $+7$; in $\mathrm{MnO_2}$, Mn is $+4$.

Step 2 — Oxidation of thiosulphate.

In $\mathrm{S_2O_3^{2-}}$ the average oxidation state of S is $+2$ (since $2x + 3(-2) = -2$). In $\mathrm{SO_4^{2-}}$, S is $+6$. Each $\mathrm{S_2O_3^{2-}}$ produces two sulphate ions:

Step 3 — Equalise the electrons.

LCM of 3 and 8 is 24, so multiply the reduction by 8 and the oxidation by 3:

Step 4 — Complete with water/hydroxide (charge and atom balance). …

Step 1 — Identify the quantum numbers.

For a $3p$ orbital:

• principal quantum number $n = 3$ (from the digit),
• azimuthal quantum number $l = 1$ (since $p \Rightarrow l = 1$; $s=0$, $p=1$, $d=2$, $f=3$).

Step 2 — Angular (nodal-plane) nodes.

These come from the angular part $Y_{l,m}(\theta,\phi)$ of the wavefunction, and their count is exactly

Physically: a $p$ orbital is dumbbell-shaped with one nodal plane through the nucleus (e.g. the $xy$-plane for $p_z$). Note this is independent of $n$ — $2p$, $3p$, $4p$ all have exactly 1 angular node.

Step 3 — Radial (spherical) nodes.

These come from the radial part $R_{n,l}(r)$, and

$\text{radial nodes} = n - l - 1 = 3 - 1 - 1 = \boxed{1}$ …

Step 1 — The integrated first-order rate law.

Starting from $-\dfrac{d[A]}{dt} = k[A]$ and integrating,

(The factor $2.303$ converts $\ln$ to $\log_{10}$.)

Step 2 — Translate "99% completion".

If 99% of A has reacted, the fraction remaining is $1\%$:

This is the step candidates most often get wrong — the ratio uses what is left, not what has gone.

Step 3 — Substitute.

Step 4 — Cross-check with half-lives. …

Step 1 — The first-order integrated rate law.

where $a$ is the initial amount and $x$ the amount reacted. The key property of first order: $k$ depends only on the ratio $a/(a-x)$, not on the absolute concentration.

Step 2 — Use the given data to get $k$.

60% complete ⇒ 40% remains ⇒ $\dfrac{a}{a-x}=\dfrac{100}{40}=2.5$, with $t=50$ min:

Step 3 — Set up the second condition.

93.6% complete ⇒ remaining $=100-93.6=6.4\%$ ⇒

Step 4 — Spot the elegant relation (no calculator needed).

Equivalently the surviving fraction $0.064=(0.4)^3$. Taking logs:

Step 5 — Get the time. …