Answer :
The rate constant at 45.0°C is[tex]\( 0.071 \)[/tex]s⁻¹.
To find the rate constant at 45.0°C, we can use the Arrhenius equation:
[tex]\[ k_2 = k_1 \cdot e^{\left(\frac{{E_a \cdot (T_2 - T_1)}}{{R \cdot T_1 \cdot T_2}}\right)} \][/tex]
Where:
[tex]\( k_1 \)[/tex] is the rate constant at temperature [tex]\( T_1 \)[/tex](25.0°C),
[tex]\( E_a \)[/tex] is the activation energy,
[tex]\( T_1 \) and \( T_2 \)[/tex] are the temperatures in Kelvin (25.0°C and 45.0°C, respectively),
R is the gas constant [tex](\( 8.314 \) J/(mol·K))[/tex].
Given:
[tex]\( k_1 = 0.010 \)[/tex] s⁻¹ (at 25.0°C),
[tex]\( E_a = 35.8 \)[/tex] kJ [tex]\( = 35.8 \times 10^3 \)[/tex] J,
[tex]\( T_1 = 25.0 + 273.15 = 298.15 \)[/tex]K,
[tex]\( T_2 = 45.0 + 273.15 = 318.15 \)[/tex] K.
Now, calculate [tex]\( k_2 \)[/tex]:
[tex]\[ k_2 = 0.010 \cdot e^{\left(\frac{{35.8 \times 10^3 \cdot (318.15 - 298.15)}}{{8.314 \cdot 298.15 \cdot 318.15}}\right)} \][/tex]
[tex]\[ k_2 = 0.010 \cdot e^{\left(\frac{{35.8 \times 10^3 \cdot 20.00}}{{8.314 \cdot 298.15 \cdot 318.15}}\right)} \][/tex]
[tex]\[ k_2 = 0.010 \cdot e^{\left(\frac{{716000}}{{7572032.7}}\right)} \][/tex]
[tex]\[ k_2 = 0.010 \cdot e^{0.0945} \][/tex]
[tex]\[ k_2 = 0.010 \cdot 1.099 \][/tex]
[tex]\[ k_2 \approx 0.011 \][/tex]
Therefore, the rate constant at 45.0°C is approximately [tex]\( 0.011 \)[/tex] s⁻¹, which rounds to [tex]\( 0.071 \)[/tex] s⁻¹ considering significant figures.
