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The relationship between Kp and Kc is Kp...

The relationship between `K_p and K_c` is `K_p=K_c(RT)^(Deltan)` . What would be the value of `Deltan` for the reaction :
`NH_4Cl(s) hArr NH_3(g) +HCl(g)` ?

A

1

B

0.5

C

1.5

D

2

Text Solution

AI Generated Solution

The correct Answer is:
To determine the value of Δn for the reaction: \[ NH_4Cl(s) \rightleftharpoons NH_3(g) + HCl(g) \] we need to follow these steps: ### Step 1: Identify the reactants and products In the given reaction, we have: - Reactant: \( NH_4Cl(s) \) (solid) - Products: \( NH_3(g) \) (gas) and \( HCl(g) \) (gas) ### Step 2: Count the moles of gaseous products Next, we count the total number of moles of gaseous products: - From the products, we have: - 1 mole of \( NH_3(g) \) - 1 mole of \( HCl(g) \) Thus, the total moles of gaseous products = 1 + 1 = 2 moles. ### Step 3: Count the moles of gaseous reactants Now, we check the moles of gaseous reactants: - The reactant \( NH_4Cl(s) \) is a solid and does not contribute to the gaseous moles. Thus, the total moles of gaseous reactants = 0 moles. ### Step 4: Calculate Δn Now, we can calculate Δn using the formula: \[ \Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \] Substituting the values we found: \[ \Delta n = 2 - 0 = 2 \] ### Final Answer The value of Δn for the reaction is: \[ \Delta n = 2 \] ---
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The relation between K_(p) and K_(c) is K_(p)=K_(c)(RT)^(Deltan) unit of K_(p)=(atm)^(Deltan) , unit of K_(c)=(mol L^(-1))^(Deltan) H_(3)ClO_(4) is a tribasic acid, it undergoes ionisation as H_(3)ClO_(4) hArr H^(o+)+H_(2)ClO_(4)^(-), K_(1) H_(2)ClO_(4)^(-) hArr H^(o+)+HClO_(4)^(2-), K_(2) HClO_(4)^(2-) hArr H^(o+)+ClO_(4)^(3-), K_(3) Then, equilibrium constant for the following reaction will be: H_(3)ClO_(4) hArr 3H^(o+)+ClO_(4)^(3-)

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