For silver, Cp (JK–1 mol–1) = 23 + 0.01 T. If the temperature (T) of 3 moles of silver is raised from 300K to 1000 K at 1 atm pressure, the value of DH will be close to:

Engineering

Chemistry

Enthalpy and Internal Energy Relation

Enthalpy Calculation and Meaning

First Law and Various Process

Question

For silver, C_p (JK^–1 mol^–1) = 23 + 0.01 T. If the temperature (T) of 3 moles of silver is raised from 300K to 1000 K at 1 atm pressure, the value of DH will be close to:

13 kJ

16 kJ

62 kJ

21 kJ

$ΔH = n \int_{T_{1}}^{T_{2}} C_{p . m} dT = 3 \times \int_{300}^{1000} (23 + 0 .01 T) dT$

$= 3 [23 (1000 - 300)] + \frac{0 .01}{2} [{(1000)}^{2} - {(300)}^{2}]$

= 61950 J ≈ 62 KJ

To find the enthalpy change (ΔH) for heating silver, we use the formula for enthalpy change at constant pressure, which is given by the integral of the heat capacity at constant pressure (C_p) with respect to temperature (T). The provided C_p is a function of T: C_p = 23 + 0.01T J·K^–1·mol^–1.

Step 1: Write the expression for ΔH for n moles. $Δ H = n \int_{T_{1}}^{T_{2}} C_{p} d T$

Step 2: Substitute the given values: n = 3 mol, T₁ = 300 K, T₂ = 1000 K, and C_p = 23 + 0.01T. $Δ H = 3 \int_{300}^{1000} (23 + 0.01 T) d T$

Step 3: Integrate the expression. $Δ H = 3 (\int_{300}^{1000} 23 d T + \int_{300}^{1000} 0.01 T d T) = 3 (23 (T) |_{300}^{1000} + 0.01 \frac{T^{2}}{2} |_{300}^{1000})$

Step 4: Evaluate the definite integrals. $23 (1000 - 300) = 23 \times 700 = 16100$
$\frac{0.01}{2} (1000^{2} - 300^{2}) = 0.005 (1000000 - 90000) = 0.005 \times 910000 = 4550$

Step 5: Sum the integrals and multiply by n. $Δ H = 3 \times (16100 + 4550) = 3 \times 20650 = 61950 J$

Step 6: Convert to kJ. $Δ H = 61950 / 1000 = 61.95 kJ$ , which is close to 62 kJ.

Final Answer: 62 kJ

Detailed Solution

Related Topics and Formulae

Detailed Solution

Related Topics and Formulae

Student who searched for similar questions

Questions 1

Questions 2

Questions 3

Questions 4

Questions 5