Which one of the following has the same number of atoms as are in 6g of H2O
0.4G He
22g CO2
1g H2
12g CO
To solve the problem, we need to find which option contains the same number of atoms as 6 g of H2O.
1. Calculate Moles of H2O:
Molar mass of water (H2O) = 18 g/mol
Moles of water = $ \frac{6}{18} = 0.333 \, \text{mol} $
2. Calculate Number of Atoms in H2O:
Each molecule of water contains 3 atoms (2 Hydrogen + 1 Oxygen).
So, total atoms = $0.333 \times 6.022 \times 10^{23} \times 3$
= $6.022 \times 10^{23} \times 1 = 6.022 \times 10^{23}$ atoms
3. Find Which Option Has the Same Number of Atoms:
Final Answer:
The option with the same number of atoms as in 6 g of water is 1 g H2.
For the reaction A(g) $\rightleftharpoons$ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.
[Given: R = 0.0831 atm $mol^{–1} K^{–1}$]
$K_p$ for the reaction at 1000 K is:
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]
Consider the following gas phase dissociation, PCl$_5$(g) $\rightleftharpoons$ PCl$_3$(g) + Cl$_2$(g) with equilibrium constant K$_p$ at a particular temperature and at pressure P. The degree of dissociation ($\alpha$) for PCl$_5$(g) is
PCl$_5$(g) $\rightleftharpoons$ PCl$_3$(g) + Cl$_2$(g)
Law of Chemical Equilibrium states that at a constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants each raised to a power equal to the corresponding stoichiometric coefficients as represented by the balanced chemical equation.
Let us consider a general reversible reaction;
A+B ↔ C+D
After some time, there is a reduction in reactants A and B and an accumulation of the products C and D. As a result, the rate of the forward reaction decreases and that of backward reaction increases.
Eventually, the two reactions occur at the same rate and a state of equilibrium is attained.
By applying the Law of Mass Action;
The rate of forward reaction;
Rf = Kf [A]a [B]b
The rate of backward reaction;
Rb = Kb [C]c [D]d
Where,
[A], [B], [C] and [D] are the concentrations of A, B, C and D at equilibrium respectively.
a, b, c, and d are the stoichiometric coefficients of A, B, C and D respectively.
Kf and Kb are the rate constants of forward and backward reactions.
However, at equilibrium,
Rate of forward reaction = Rate of backward reaction.
Kc is called the equilibrium constant expressed in terms of molar concentrations.
The above equation is known as the equation of Law of Chemical Equilibrium.