CINCINNATI — Cincinnati Bengals quarterback Joe Burrow was named AFC Offensive Player of the Week for his performance in Sunday’s win at the San Francisco 49ers.
He completed 28 of 32 passes for 283 yards and three touchdowns without throwing an interception in a 31-17 win on Sunday.
His career-high 87.5 completion percentage led all NFL passers in Week 8 and was the second-highest in a game in team history.
Burrow completed 19 consecutive pass attempts in the first half, the most by most in a single game by any NFL quarterback this season.
He also threw his 90th career touchdown pass in his 49th career game, which is tied as the sixth-fastest quarterback in NFL history to pass for 90 touchdowns.
This is Burrow’s sixth career AFC Offensive Player of the Week Award.
The Bengals host the Buffalo Bills on Sunday Night Football on November 5.
Kickoff is scheduled for 8:20 p.m. at Paycor Stadium.
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